如何实现二叉树？

Question

在 Python 中可用于实现二叉树的最佳数据结构是什么？

Answer 1

这是我对二叉搜索树的简单递归实现。

#!/usr/bin/python

class Node:
    def __init__(self, val):
        self.l = None
        self.r = None
        self.v = val

class Tree:
    def __init__(self):
        self.root = None

    def getRoot(self):
        return self.root

    def add(self, val):
        if self.root is None:
            self.root = Node(val)
        else:
            self._add(val, self.root)

    def _add(self, val, node):
        if val < node.v:
            if node.l is not None:
                self._add(val, node.l)
            else:
                node.l = Node(val)
        else:
            if node.r is not None:
                self._add(val, node.r)
            else:
                node.r = Node(val)

    def find(self, val):
        if self.root is not None:
            return self._find(val, self.root)
        else:
            return None

    def _find(self, val, node):
        if val == node.v:
            return node
        elif (val < node.v and node.l is not None):
            return self._find(val, node.l)
        elif (val > node.v and node.r is not None):
            return self._find(val, node.r)

    def deleteTree(self):
        # garbage collector will do this for us. 
        self.root = None

    def printTree(self):
        if self.root is not None:
            self._printTree(self.root)

    def _printTree(self, node):
        if node is not None:
            self._printTree(node.l)
            print(str(node.v) + ' ')
            self._printTree(node.r)

#     3
# 0     4
#   2      8
tree = Tree()
tree.add(3)
tree.add(4)
tree.add(0)
tree.add(8)
tree.add(2)
tree.printTree()
print(tree.find(3).v)
print(tree.find(10))
tree.deleteTree()
tree.printTree()

Answer 2

[面试需要什么] Node 类是足够的数据结构来表示二叉树。

（虽然其他答案大多是正确的，但二叉树不需要它们：无需扩展对象类，无需成为 BST，无需导入双端队列）。

class Node:

    def __init__(self, value = None):
        self.left  = None
        self.right = None
        self.value = value

这是一个树的例子：

n1 = Node(1)
n2 = Node(2)
n3 = Node(3)
n1.left  = n2
n1.right = n3

在此示例中，n1 是具有 n2、n3 作为其子级的树的根。

Answer 3

# simple binary tree
# in this implementation, a node is inserted between an existing node and the root


class BinaryTree():

    def __init__(self,rootid):
      self.left = None
      self.right = None
      self.rootid = rootid

    def getLeftChild(self):
        return self.left
    def getRightChild(self):
        return self.right
    def setNodeValue(self,value):
        self.rootid = value
    def getNodeValue(self):
        return self.rootid

    def insertRight(self,newNode):
        if self.right == None:
            self.right = BinaryTree(newNode)
        else:
            tree = BinaryTree(newNode)
            tree.right = self.right
            self.right = tree

    def insertLeft(self,newNode):
        if self.left == None:
            self.left = BinaryTree(newNode)
        else:
            tree = BinaryTree(newNode)
            tree.left = self.left
            self.left = tree


def printTree(tree):
        if tree != None:
            printTree(tree.getLeftChild())
            print(tree.getNodeValue())
            printTree(tree.getRightChild())



# test tree

def testTree():
    myTree = BinaryTree("Maud")
    myTree.insertLeft("Bob")
    myTree.insertRight("Tony")
    myTree.insertRight("Steven")
    printTree(myTree)

在此处了解更多信息：-这是一个非常简单的二叉树implementation。

This 是一个很好的教程，中间有问题

Answer 4

我不禁注意到这里的大多数答案都是实现二叉搜索树。二叉搜索树！= 二叉树。

• 二叉搜索树有一个非常特殊的属性：对于任何节点 X，X 的键都大于其左孩子的任何后代的键，并且小于其右孩子的任何后代的键。

• 二叉树没有这种限制。二叉树是一个简单的数据结构，包含一个“键”元素和两个子元素，即“左”和“右”。

• 树是二叉树的更一般情况，其中每个节点可以有任意数量的子节点。通常，每个节点都有一个“子”元素，其类型为列表/数组。

现在，为了回答 OP 的问题，我在 Python 中包含了二叉树的完整实现。存储每个 BinaryTreeNode 的底层数据结构是一个字典，因为它提供了最佳的 O(1) 查找。我还实现了深度优先和广度优先遍历。这些是在树上执行的非常常见的操作。

from collections import deque

class BinaryTreeNode:
    def __init__(self, key, left=None, right=None):
        self.key = key
        self.left = left
        self.right = right

    def __repr__(self):
        return "%s l: (%s) r: (%s)" % (self.key, self.left, self.right)

    def __eq__(self, other):
        if self.key == other.key and \
            self.right == other.right and \
                self.left == other.left:
            return True
        else:
            return False

class BinaryTree:
    def __init__(self, root_key=None):
        # maps from BinaryTreeNode key to BinaryTreeNode instance.
        # Thus, BinaryTreeNode keys must be unique.
        self.nodes = {}
        if root_key is not None:
            # create a root BinaryTreeNode
            self.root = BinaryTreeNode(root_key)
            self.nodes[root_key] = self.root

    def add(self, key, left_key=None, right_key=None):
        if key not in self.nodes:
            # BinaryTreeNode with given key does not exist, create it
            self.nodes[key] = BinaryTreeNode(key)
        # invariant: self.nodes[key] exists

        # handle left child
        if left_key is None:
            self.nodes[key].left = None
        else:
            if left_key not in self.nodes:
                self.nodes[left_key] = BinaryTreeNode(left_key)
            # invariant: self.nodes[left_key] exists
            self.nodes[key].left = self.nodes[left_key]

        # handle right child
        if right_key == None:
            self.nodes[key].right = None
        else:
            if right_key not in self.nodes:
                self.nodes[right_key] = BinaryTreeNode(right_key)
            # invariant: self.nodes[right_key] exists
            self.nodes[key].right = self.nodes[right_key]

    def remove(self, key):
        if key not in self.nodes:
            raise ValueError('%s not in tree' % key)
        # remove key from the list of nodes
        del self.nodes[key]
        # if node removed is left/right child, update parent node
        for k in self.nodes:
            if self.nodes[k].left and self.nodes[k].left.key == key:
                self.nodes[k].left = None
            if self.nodes[k].right and self.nodes[k].right.key == key:
                self.nodes[k].right = None
        return True

    def _height(self, node):
        if node is None:
            return 0
        else:
            return 1 + max(self._height(node.left), self._height(node.right))

    def height(self):
        return self._height(self.root)

    def size(self):
        return len(self.nodes)

    def __repr__(self):
        return str(self.traverse_inorder(self.root))

    def bfs(self, node):
        if not node or node not in self.nodes:
            return
        reachable = []    
        q = deque()
        # add starting node to queue
        q.append(node)
        while len(q):
            visit = q.popleft()
            # add currently visited BinaryTreeNode to list
            reachable.append(visit)
            # add left/right children as needed
            if visit.left:
                q.append(visit.left)
            if visit.right:
                q.append(visit.right)
        return reachable

    # visit left child, root, then right child
    def traverse_inorder(self, node, reachable=None):
        if not node or node.key not in self.nodes:
            return
        if reachable is None:
            reachable = []
        self.traverse_inorder(node.left, reachable)
        reachable.append(node.key)
        self.traverse_inorder(node.right, reachable)
        return reachable

    # visit left and right children, then root
    # root of tree is always last to be visited
    def traverse_postorder(self, node, reachable=None):
        if not node or node.key not in self.nodes:
            return
        if reachable is None:
            reachable = []
        self.traverse_postorder(node.left, reachable)
        self.traverse_postorder(node.right, reachable)
        reachable.append(node.key)
        return reachable

    # visit root, left, then right children
    # root is always visited first
    def traverse_preorder(self, node, reachable=None):
        if not node or node.key not in self.nodes:
            return
        if reachable is None:
            reachable = []
        reachable.append(node.key)
        self.traverse_preorder(node.left, reachable)
        self.traverse_preorder(node.right, reachable)
        return reachable

Answer 5

BST在Python中的简单实现

class TreeNode:
    def __init__(self, value):
        self.left = None
        self.right = None
        self.data = value

class Tree:
    def __init__(self):
        self.root = None

    def addNode(self, node, value):
        if(node==None):
            self.root = TreeNode(value)
        else:
            if(value<node.data):
                if(node.left==None):
                    node.left = TreeNode(value)
                else:
                    self.addNode(node.left, value)
            else:
                if(node.right==None):
                    node.right = TreeNode(value)
                else:
                    self.addNode(node.right, value)

    def printInorder(self, node):
        if(node!=None):
            self.printInorder(node.left)
            print(node.data)
            self.printInorder(node.right)

def main():
    testTree = Tree()
    testTree.addNode(testTree.root, 200)
    testTree.addNode(testTree.root, 300)
    testTree.addNode(testTree.root, 100)
    testTree.addNode(testTree.root, 30)
    testTree.printInorder(testTree.root)

Answer 6

使用列表实现二叉树的一种非常快速且简单的方法。 不是最有效的，也不能很好地处理 nil 值。 但它非常透明（至少对我而言）：

def _add(node, v):
    new = [v, [], []]
    if node:
        left, right = node[1:]
        if not left:
            left.extend(new)
        elif not right:
            right.extend(new)
        else:
            _add(left, v)
    else:
        node.extend(new)

def binary_tree(s):
    root = []
    for e in s:
        _add(root, e)
    return root

def traverse(n, order):
    if n:
        v = n[0]
        if order == 'pre':
            yield v
        for left in traverse(n[1], order):
            yield left
        if order == 'in':
            yield v
        for right in traverse(n[2], order):
            yield right
        if order == 'post':
            yield v

从一个可迭代对象构造一棵树：

 >>> tree = binary_tree('A B C D E'.split())
 >>> print tree
 ['A', ['B', ['D', [], []], ['E', [], []]], ['C', [], []]]

遍历一棵树：

 >>> list(traverse(tree, 'pre')), list(traverse(tree, 'in')), list(traverse(tree, 'post'))
 (['A', 'B', 'D', 'E', 'C'],
  ['D', 'B', 'E', 'A', 'C'],
  ['D', 'E', 'B', 'C', 'A'])

Answer 7

你不需要有两个类

class Tree:
    val = None
    left = None
    right = None

    def __init__(self, val):
        self.val = val


    def insert(self, val):
        if self.val is not None:
            if val < self.val:
                if self.left is not None:
                    self.left.insert(val)
                else:
                    self.left = Tree(val)
            elif val > self.val:
                if self.right is not None:
                    self.right.insert(val)
                else:
                    self.right = Tree(val)
            else:
                return
        else:
            self.val = val
            print("new node added")

    def showTree(self):
        if self.left is not None:
            self.left.showTree()
        print(self.val, end = ' ')
        if self.right is not None:
            self.right.showTree()

Answer 8

基于Node 的连接节点类是一种标准方法。这些可能很难想象。

灵感来自于 Python 模式 - 实现图上的 essay，考虑一个简单的字典：

给定

二叉树

               a
              / \
             b   c
            / \   \
           d   e   f

代码

制作一个唯一节点的字典：

tree = {
   "a": ["b", "c"],
   "b": ["d", "e"],
   "c": [None, "f"],
   "d": [None, None],
   "e": [None, None],
   "f": [None, None],
}

---

详情

• 每个键值对都是一个唯一节点，指向其子节点。
• 一个列表（或元组）包含一对有序的左/右孩子。
• 对于 有序插入的字典，假设第一个条目是根。
• 常用方法可以是改变或遍历字典的函数（参见find_all_paths()）。

基于树的函数通常包括以下常见操作：

• 遍历：以给定的顺序产生每个节点（通常是从左到右）
  • 广度优先搜索 (BFS)：遍历关卡
  • 深度优先搜索 (DFS)：首先遍历分支（前/中/后顺序）
• 插入：根据孩子的数量向树中添加一个节点
• remove：根据子节点的数量删除一个节点
• 更新：将缺失的节点从一棵树合并到另一棵树
• visit：产生遍历节点的值

尝试实现所有这些操作。 在这里，我们演示了这些函数中的一个 - BFS 遍历：

示例

import collections as ct


def traverse(tree):
    """Yield nodes from a tree via BFS."""
    q = ct.deque()                                         # 1
    root = next(iter(tree))                                # 2
    q.append(root)
    
    while q:
        node = q.popleft()
        children = filter(None, tree.get(node))
        for n in children:                                 # 3 
            q.append(n)
        yield node

list(traverse(tree))
# ['a', 'b', 'c', 'd', 'e', 'f']

这是一个breadth-first search (level-order) algorithm 应用于节点和子节点的字典。

1. 初始化FIFO queue。我们使用deque，但queue 或list 有效（后者效率低下）。
2. 获取根节点并将其排入队列（假设根节点是字典中的第一个条目，Python 3.6+）。
3. 迭代地使节点出队，将其子节点入队并产生节点值。

另请参阅树上的这个深入的tutorial。

---

洞察力

一般来说，遍历的好处是，我们可以轻松地将后一种迭代方法更改为 depth-first search (DFS)，只需将队列替换为 stack（也称为 LIFO 队列）。这仅仅意味着我们从入队的同一侧出队。 DFS 允许我们搜索每个分支。

怎么样？由于我们使用的是deque，我们可以通过将node = q.popleft() 更改为node = q.pop()（右）来模拟堆栈。结果是正确的，pre-ordered DFS: ['a', 'c', 'f', 'b', 'e', 'd']。

Answer 9

更“Pythonic”？

class Node:
    def __init__(self, value):
        self.value = value
        self.left = None
        self.right = None

    def __repr__(self):
        return str(self.value)



class BST:
    def __init__(self):
        self.root = None

    def __repr__(self):
        self.sorted = []
        self.get_inorder(self.root)
        return str(self.sorted)

    def get_inorder(self, node):
        if node:
            self.get_inorder(node.left)
            self.sorted.append(str(node.value))
            self.get_inorder(node.right)

    def add(self, value):
        if not self.root:
            self.root = Node(value)
        else:
            self._add(self.root, value)

    def _add(self, node, value):
        if value <= node.value:
            if node.left:
                self._add(node.left, value)
            else:
                node.left = Node(value)
        else:
            if node.right:
                self._add(node.right, value)
            else:
                node.right = Node(value)



from random import randint

bst = BST()

for i in range(100):
    bst.add(randint(1, 1000))
print (bst)

Answer 10

#!/usr/bin/python

class BinaryTree:
    def __init__(self, left, right, data):
        self.left = left
        self.right = right
        self.data = data


    def pre_order_traversal(root):
        print(root.data, end=' ')

        if root.left != None:
            pre_order_traversal(root.left)

        if root.right != None:
            pre_order_traversal(root.right)

    def in_order_traversal(root):
        if root.left != None:
            in_order_traversal(root.left)
        print(root.data, end=' ')
        if root.right != None:
            in_order_traversal(root.right)

    def post_order_traversal(root):
        if root.left != None:
            post_order_traversal(root.left)
        if root.right != None:
            post_order_traversal(root.right)
        print(root.data, end=' ')

Answer 11

import random

class TreeNode:
    def __init__(self, key):
        self.key = key
        self.left = None
        self.right = None
        self.p = None

class BinaryTree:
    def __init__(self):
        self.root = None

    def length(self):
        return self.size

    def inorder(self, node):
        if node == None:
            return None
        else:
            self.inorder(node.left)
            print node.key,
            self.inorder(node.right)

    def search(self, k):
        node = self.root
        while node != None:
            if node.key == k:
                return node
            if node.key > k:
                node = node.left
            else:
                node = node.right
        return None

    def minimum(self, node):
        x = None
        while node.left != None:
            x = node.left
            node = node.left
        return x

    def maximum(self, node):
        x = None
        while node.right != None:
            x = node.right
            node = node.right
        return x

    def successor(self, node):
        parent = None
        if node.right != None:
            return self.minimum(node.right)
        parent = node.p
        while parent != None and node == parent.right:
            node = parent
            parent = parent.p
        return parent

    def predecessor(self, node):
        parent = None
        if node.left != None:
            return self.maximum(node.left)
        parent = node.p
        while parent != None and node == parent.left:
            node = parent
            parent = parent.p
        return parent

    def insert(self, k):
        t = TreeNode(k)
        parent = None
        node = self.root
        while node != None:
            parent = node
            if node.key > t.key:
                node = node.left
            else:
                node = node.right
        t.p = parent
        if parent == None:
            self.root = t
        elif t.key < parent.key:
            parent.left = t
        else:
            parent.right = t
        return t


    def delete(self, node):
        if node.left == None:
            self.transplant(node, node.right)
        elif node.right == None:
            self.transplant(node, node.left)
        else:
            succ = self.minimum(node.right)
            if succ.p != node:
                self.transplant(succ, succ.right)
                succ.right = node.right
                succ.right.p = succ
            self.transplant(node, succ)
            succ.left = node.left
            succ.left.p = succ

    def transplant(self, node, newnode):
        if node.p == None:
            self.root = newnode
        elif node == node.p.left:
            node.p.left = newnode
        else:
            node.p.right = newnode
        if newnode != None:
            newnode.p = node.p

Answer 12

此实现支持插入、查找和删除操作，而不会破坏树的结构。这不是一棵平衡树。

# Class for construct the nodes of the tree. (Subtrees)
class Node:
def __init__(self, key, parent_node = None):
    self.left = None
    self.right = None
    self.key = key
    if parent_node == None:
        self.parent = self
    else:
        self.parent = parent_node

# Class with the  structure of the tree. 
# This Tree is not balanced.
class Tree:
def __init__(self):
    self.root = None

# Insert a single element
def insert(self, x):
    if(self.root == None):
        self.root = Node(x)
    else:
        self._insert(x, self.root)

def _insert(self, x, node):
    if(x < node.key):
        if(node.left == None):
            node.left = Node(x, node)
        else:
            self._insert(x, node.left)
    else:
        if(node.right == None):
            node.right = Node(x, node)
        else:
            self._insert(x, node.right)

# Given a element, return a node in the tree with key x. 
def find(self, x):
    if(self.root == None):
        return None
    else:
        return self._find(x, self.root)
def _find(self, x, node):
    if(x == node.key):
        return node
    elif(x < node.key):
        if(node.left == None):
            return None
        else:
            return self._find(x, node.left)
    elif(x > node.key):
        if(node.right == None):
            return None
        else:
            return self._find(x, node.right)

# Given a node, return the node in the tree with the next largest element.
def next(self, node):
    if node.right != None:
        return self._left_descendant(node.right)
    else:
        return self._right_ancestor(node)

def _left_descendant(self, node):
    if node.left == None:
        return node
    else:
        return self._left_descendant(node.left)

def _right_ancestor(self, node):
    if node.key <= node.parent.key:
        return node.parent
    else:
        return self._right_ancestor(node.parent)

# Delete an element of the tree
def delete(self, x):
    node = self.find(x)
    if node == None:
        print(x, "isn't in the tree")
    else:
        if node.right == None:
            if node.left == None:
                if node.key < node.parent.key:
                    node.parent.left = None
                    del node # Clean garbage
                else:
                    node.parent.right = None
                    del Node # Clean garbage
            else:
                node.key = node.left.key
                node.left = None
        else:
            x = self.next(node)
            node.key = x.key
            x = None


# tests
t = Tree()
t.insert(5)
t.insert(8)
t.insert(3)
t.insert(4)
t.insert(6)
t.insert(2)

t.delete(8)
t.delete(5)

t.insert(9)
t.insert(1)

t.delete(2)
t.delete(100)

# Remember: Find method return the node object. 
# To return a number use t.find(nº).key
# But it will cause an error if the number is not in the tree.
print(t.find(5)) 
print(t.find(8))
print(t.find(4))
print(t.find(6))
print(t.find(9))

Answer 13

一个很好的二叉搜索树实现，取自here：

'''
A binary search Tree
'''
from __future__ import print_function
class Node:

    def __init__(self, label, parent):
        self.label = label
        self.left = None
        self.right = None
        #Added in order to delete a node easier
        self.parent = parent

    def getLabel(self):
        return self.label

    def setLabel(self, label):
        self.label = label

    def getLeft(self):
        return self.left

    def setLeft(self, left):
        self.left = left

    def getRight(self):
        return self.right

    def setRight(self, right):
        self.right = right

    def getParent(self):
        return self.parent

    def setParent(self, parent):
        self.parent = parent

class BinarySearchTree:

    def __init__(self):
        self.root = None

    def insert(self, label):
        # Create a new Node
        new_node = Node(label, None)
        # If Tree is empty
        if self.empty():
            self.root = new_node
        else:
            #If Tree is not empty
            curr_node = self.root
            #While we don't get to a leaf
            while curr_node is not None:
                #We keep reference of the parent node
                parent_node = curr_node
                #If node label is less than current node
                if new_node.getLabel() < curr_node.getLabel():
                #We go left
                    curr_node = curr_node.getLeft()
                else:
                    #Else we go right
                    curr_node = curr_node.getRight()
            #We insert the new node in a leaf
            if new_node.getLabel() < parent_node.getLabel():
                parent_node.setLeft(new_node)
            else:
                parent_node.setRight(new_node)
            #Set parent to the new node
            new_node.setParent(parent_node)      

    def delete(self, label):
        if (not self.empty()):
            #Look for the node with that label
            node = self.getNode(label)
            #If the node exists
            if(node is not None):
                #If it has no children
                if(node.getLeft() is None and node.getRight() is None):
                    self.__reassignNodes(node, None)
                    node = None
                #Has only right children
                elif(node.getLeft() is None and node.getRight() is not None):
                    self.__reassignNodes(node, node.getRight())
                #Has only left children
                elif(node.getLeft() is not None and node.getRight() is None):
                    self.__reassignNodes(node, node.getLeft())
                #Has two children
                else:
                    #Gets the max value of the left branch
                    tmpNode = self.getMax(node.getLeft())
                    #Deletes the tmpNode
                    self.delete(tmpNode.getLabel())
                    #Assigns the value to the node to delete and keesp tree structure
                    node.setLabel(tmpNode.getLabel())

    def getNode(self, label):
        curr_node = None
        #If the tree is not empty
        if(not self.empty()):
            #Get tree root
            curr_node = self.getRoot()
            #While we don't find the node we look for
            #I am using lazy evaluation here to avoid NoneType Attribute error
            while curr_node is not None and curr_node.getLabel() is not label:
                #If node label is less than current node
                if label < curr_node.getLabel():
                    #We go left
                    curr_node = curr_node.getLeft()
                else:
                    #Else we go right
                    curr_node = curr_node.getRight()
        return curr_node

    def getMax(self, root = None):
        if(root is not None):
            curr_node = root
        else:
            #We go deep on the right branch
            curr_node = self.getRoot()
        if(not self.empty()):
            while(curr_node.getRight() is not None):
                curr_node = curr_node.getRight()
        return curr_node

    def getMin(self, root = None):
        if(root is not None):
            curr_node = root
        else:
            #We go deep on the left branch
            curr_node = self.getRoot()
        if(not self.empty()):
            curr_node = self.getRoot()
            while(curr_node.getLeft() is not None):
                curr_node = curr_node.getLeft()
        return curr_node

    def empty(self):
        if self.root is None:
            return True
        return False

    def __InOrderTraversal(self, curr_node):
        nodeList = []
        if curr_node is not None:
            nodeList.insert(0, curr_node)
            nodeList = nodeList + self.__InOrderTraversal(curr_node.getLeft())
            nodeList = nodeList + self.__InOrderTraversal(curr_node.getRight())
        return nodeList

    def getRoot(self):
        return self.root

    def __isRightChildren(self, node):
        if(node == node.getParent().getRight()):
            return True
        return False

    def __reassignNodes(self, node, newChildren):
        if(newChildren is not None):
            newChildren.setParent(node.getParent())
        if(node.getParent() is not None):
            #If it is the Right Children
            if(self.__isRightChildren(node)):
                node.getParent().setRight(newChildren)
            else:
                #Else it is the left children
                node.getParent().setLeft(newChildren)

    #This function traversal the tree. By default it returns an
    #In order traversal list. You can pass a function to traversal
    #The tree as needed by client code
    def traversalTree(self, traversalFunction = None, root = None):
        if(traversalFunction is None):
            #Returns a list of nodes in preOrder by default
            return self.__InOrderTraversal(self.root)
        else:
            #Returns a list of nodes in the order that the users wants to
            return traversalFunction(self.root)

    #Returns an string of all the nodes labels in the list 
    #In Order Traversal
    def __str__(self):
        list = self.__InOrderTraversal(self.root)
        str = ""
        for x in list:
            str = str + " " + x.getLabel().__str__()
        return str

def InPreOrder(curr_node):
    nodeList = []
    if curr_node is not None:
        nodeList = nodeList + InPreOrder(curr_node.getLeft())
        nodeList.insert(0, curr_node.getLabel())
        nodeList = nodeList + InPreOrder(curr_node.getRight())
    return nodeList

def testBinarySearchTree():
    r'''
    Example
                  8
                 / \
                3   10
               / \    \
              1   6    14
                 / \   /
                4   7 13 
    '''

    r'''
    Example After Deletion
                  7
                 / \
                1   4

    '''
    t = BinarySearchTree()
    t.insert(8)
    t.insert(3)
    t.insert(6)
    t.insert(1)
    t.insert(10)
    t.insert(14)
    t.insert(13)
    t.insert(4)
    t.insert(7)

    #Prints all the elements of the list in order traversal
    print(t.__str__())

    if(t.getNode(6) is not None):
        print("The label 6 exists")
    else:
        print("The label 6 doesn't exist")

    if(t.getNode(-1) is not None):
        print("The label -1 exists")
    else:
        print("The label -1 doesn't exist")

    if(not t.empty()):
        print(("Max Value: ", t.getMax().getLabel()))
        print(("Min Value: ", t.getMin().getLabel()))

    t.delete(13)
    t.delete(10)
    t.delete(8)
    t.delete(3)
    t.delete(6)
    t.delete(14)

    #Gets all the elements of the tree In pre order
    #And it prints them
    list = t.traversalTree(InPreOrder, t.root)
    for x in list:
        print(x)

if __name__ == "__main__":
    testBinarySearchTree()

Answer 14

我知道已经发布了许多好的解决方案，但我通常对二叉树有不同的方法：使用一些 Node 类并直接实现它更具可读性，但是当你有很多节点时，它可能会变得非常贪婪，所以我建议增加一层复杂性并将节点存储在 python 列表中，然后仅使用列表来模拟树的行为。

您仍然可以在需要时定义一个 Node 类来最终表示树中的节点，但是将它们以简单的形式 [value, left, right] 保存在列表中将占用一半或更少的内存！

这是一个将节点存储在数组中的二叉搜索树类的快速示例。它提供了基本的功能，例如添加、删除、查找...

"""
Basic Binary Search Tree class without recursion...
"""

__author__ = "@fbparis"

class Node(object):
    __slots__ = "value", "parent", "left", "right"
    def __init__(self, value, parent=None, left=None, right=None):
        self.value = value
        self.parent = parent
        self.left = left
        self.right = right

    def __repr__(self):
        return "<%s object at %s: parent=%s, left=%s, right=%s, value=%s>" % (self.__class__.__name__, hex(id(self)), self.parent, self.left, self.right, self.value)

class BinarySearchTree(object):
    __slots__ = "_tree"
    def __init__(self, *args):
        self._tree = []
        if args:
            for x in args[0]:
                self.add(x)

    def __len__(self):
        return len(self._tree)

    def __repr__(self):
        return "<%s object at %s with %d nodes>" % (self.__class__.__name__, hex(id(self)), len(self))

    def __str__(self, nodes=None, level=0):
        ret = ""
        if nodes is None:
            if len(self):
                nodes = [0]
            else:
                nodes = []
        for node in nodes:
            if node is None:
                continue
            ret += "-" * level + " %s\n" % self._tree[node][0]
            ret += self.__str__(self._tree[node][2:4], level + 1)
        if level == 0:
            ret = ret.strip()
        return ret

    def __contains__(self, value):
        if len(self):
            node_index = 0
            while self._tree[node_index][0] != value:
                if value < self._tree[node_index][0]:
                    node_index = self._tree[node_index][2]
                else:
                    node_index = self._tree[node_index][3]
                if node_index is None:
                    return False
            return True
        return False

    def __eq__(self, other):
        return self._tree == other._tree

    def add(self, value):
        if len(self):
            node_index = 0
            while self._tree[node_index][0] != value:
                if value < self._tree[node_index][0]:
                    b = self._tree[node_index][2]
                    k = 2
                else:
                    b = self._tree[node_index][3]
                    k = 3
                if b is None:
                    self._tree[node_index][k] = len(self)
                    self._tree.append([value, node_index, None, None])
                    break
                node_index = b
        else:
            self._tree.append([value, None, None, None])

    def remove(self, value):
        if len(self):
            node_index = 0
            while self._tree[node_index][0] != value:
                if value < self._tree[node_index][0]:
                    node_index = self._tree[node_index][2]
                else:
                    node_index = self._tree[node_index][3]
                if node_index is None:
                    raise KeyError
            if self._tree[node_index][2] is not None:
                b, d = 2, 3
            elif self._tree[node_index][3] is not None:
                b, d = 3, 2
            else:
                i = node_index
                b = None
            if b is not None:
                i = self._tree[node_index][b]
                while self._tree[i][d] is not None:
                    i = self._tree[i][d]
                p = self._tree[i][1]
                b = self._tree[i][b]
                if p == node_index:
                    self._tree[p][5-d] = b
                else:
                    self._tree[p][d] = b
                if b is not None:
                    self._tree[b][1] = p
                self._tree[node_index][0] = self._tree[i][0]
            else:
                p = self._tree[i][1]
                if p is not None:
                    if self._tree[p][2] == i:
                        self._tree[p][2] = None
                    else:
                        self._tree[p][3] = None
            last = self._tree.pop()
            n = len(self)
            if i < n:
                self._tree[i] = last[:]
                if last[2] is not None:
                    self._tree[last[2]][1] = i
                if last[3] is not None:
                    self._tree[last[3]][1] = i
                if self._tree[last[1]][2] == n:
                    self._tree[last[1]][2] = i
                else:
                    self._tree[last[1]][3] = i
        else:
            raise KeyError

    def find(self, value):
        if len(self):
            node_index = 0
            while self._tree[node_index][0] != value:
                if value < self._tree[node_index][0]:
                    node_index = self._tree[node_index][2]
                else:
                    node_index = self._tree[node_index][3]
                if node_index is None:
                    return None
            return Node(*self._tree[node_index])
        return None

我添加了一个 parent 属性，以便您可以删除任何节点并维护 BST 结构。

抱歉可读性，尤其是“删除”功能。基本上，当一个节点被删除时，我们弹出树数组并用最后一个元素替换它（除非我们想删除最后一个节点）。为了维护 BST 结构，移除的节点被其左子节点的最大值或右子节点的最小值替换，并且必须执行一些操作以保持索引有效但速度足够快。

我将这种技术用于更高级的东西来构建一些带有内部基数特里树的大词词典，并且我能够将内存消耗除以 7-8（您可以在此处查看示例：https://gist.github.com/fbparis/b3ddd5673b603b42c880974b23db7cda）

Answer 15

这是一个简单的解决方案，可用于构建二叉树，使用递归方法显示树，以便在以下代码中使用遍历顺序。

class Node(object):

    def __init__(self):
        self.left = None
        self.right = None
        self.value = None
    @property
    def get_value(self):
        return self.value

    @property
    def get_left(self):
        return self.left

    @property
    def get_right(self):
        return self.right

    @get_left.setter
    def set_left(self, left_node):
        self.left = left_node
    @get_value.setter
    def set_value(self, value):
        self.value = value
    @get_right.setter
    def set_right(self, right_node):
        self.right = right_node



    def create_tree(self):
        _node = Node() #creating new node.
        _x = input("Enter the node data(-1 for null)")
        if(_x == str(-1)): #for defining no child.
            return None
        _node.set_value = _x #setting the value of the node.
        print("Enter the left child of {}".format(_x))
        _node.set_left = self.create_tree() #setting the left subtree
        print("Enter the right child of {}".format(_x))
        _node.set_right = self.create_tree() #setting the right subtree.

        return _node

    def pre_order(self, root):
        if root is not None:
            print(root.get_value)
            self.pre_order(root.get_left)
            self.pre_order(root.get_right)

if __name__ == '__main__':
    node = Node()
    root_node = node.create_tree()
    node.pre_order(root_node)

代码取自：Binary Tree in Python

Answer 16

我想展示@apadana 方法的一种变体，当有相当多的节点时，它会更有用：

'''
Suppose we have the following tree
      10
    /    \
  11      9
 /  \     / \
7   12  15   8
'''
# Step 1 - Create nodes - Use a list instead of defining each node separately
nlist = [10,11,7,9,15,8,12]; n = []
for i in range(len(nlist)): n.append(Node(nlist[i]))

# Step 2 - Set each node position
n[0].left  = n[1]
n[1].left = n[2]
n[0].right = n[3]
n[3].left = n[4]
n[3].right = n[5]
n[1].right = n[6]

Answer 17

您可以在 Python 中以 OOP 方式（或构建类）制作自己的 BinaryTree 数据结构。 您可以在这里分离两个类：Node 和 BinaryTree。 “Node”类将负责为 BinaryTree 创建单个节点对象，而“BinaryTree”类是您在“Node”类之上实现二叉树所需要的。

这是我当年学习时写的代码：

class TreeNode:

    def __init__(self, data=None):
        self.data = data
        self.left = None
        self.right = None

    def __str__(self):
        return f'Node(Data={self.data}, Left={self.left}, Right={self.right})'

    def __repr__(self):
        return self.__str__()

    def get_data(self):
        return self.data

    def set_data(self, data):
        self.data = data

    def get_left(self):
        return self.left

    def set_left(self, left):
        self.left = left

    def get_right(self):
        return self.right

    def set_right(self, right):
        self.right = right

class BinaryTree:

    def __init__(self, root=None):
        self.root = TreeNode(root)

    def __str__(self):
        return f'BinaryTree({self.root})'

    def __repr__(self):
        return f'BinaryTree({self.root})'

    def insert(self, data):
        # if empty tree
        if self.root.get_data() is None:
            return self.root.set_data(data)
        new_node = TreeNode(data)
        current = self.root
        while True:
            if data < current.get_data():
                if current.get_left() is None:
                    return current.set_left(new_node)
                current = current.get_left()
                continue
            elif data > current.get_data():
                if current.get_right() is None:
                    return current.set_right(new_node)
                current = current.get_right()
                continue
            return

    # still needs other methods like the delete method, but you can
    # try it out yourself
    def delete(self, node):
        pass

def main():
    myTree = BinaryTree()
    myTree.insert(5)
    myTree.insert(3)
    myTree.insert(4)
    myTree.insert(2)
    myTree.insert(8)
    myTree.insert(9)
    myTree.insert(6)
    print(myTree)

if __name__ == '__main__':
    main()

Answer 18

按照节点的顺序，左孩子在右孩子之前。

每个节点都被认为是左右树的根节点。编写一个类来轻松创建节点：

class _Node:
    #slots are class level member,efficiently allocates memory for instance variables
    __slots__='_element','_left','_right'
    def __init__(self,element,left=None,right=None):
     # left is not a node, left is the left sub Binary Tree
     # right is the right sub Binary Tree
        self._element=element
        self._left=left
        self._right=right

这里我们编写 Binary 类：

class BinaryTree:
    def __init__(self):
        self._root=None
    def make_tree(self,e,left,right): # left=left-subtree, right=right-subtree
        # we start the tree from leaf nodes.. since it has no left and right subtrees, left and right null 
        # x.maketree(B,null,null)=[Q,B,Q] this is node x
        # y.maketree(C,null,null)=[Q,C,Q]
        # z.maketree(A,x,y)  "z" is the parent of "x" and "y"
        # each node is the root of the binary tree
        # each subtree is also considered to be Binary Tree
        self._root=_Node(e,left._root,right._root) 
    # inorder similar to infix:A+B. visit left first, then root, then right
    def inorder(self,troot):
        if troot:
            self.inorder(troot._left)
            print(troot._element,end=' ')
            self.inorder(troot._right)
   # preorder similar to prefix. +AB, visit root first,then left, then right
    def preorder(self,troot):
        if troot:
            print(troot._element,end=' ')
            self.preorder(troot._left)
            self.preorder(troot._right)
   # postorder similar to postfix. left first, then right, then root
    def postorder(self,troot):
        if troot:
            self.postorder(troot._left)
            self.postorder(troot._right)
            print(troot._element,end=' ')  
   # count the number of nodes recursively
   # recursive calls break the problem into smallest sub problems
   # we are recursively asking each node, how many children does each node have
  # if a node does not have any child, we count that node, that is why add +1. x+y+1
    def count(self,troot):
        if troot:
            x=self.count(troot._left)
            print("x",x)
            y=self.count(troot._right)
            print("y",y)
            print("x+y",x+y)
            # we add +1 because we have to count the root 
            return x+y+1
        return 0
    def height(self,troot):
        if troot:
            x=self.height(troot._left)
            y=self.height(troot._right)
            if x>y:
                return x+1
            else:
                return y+1
        return 0

现在创建二叉树：

创建6个子二叉树

x=BinaryTree()
y=BinaryTree()
z=BinaryTree()
r=BinaryTree()
s=BinaryTree()
t=BinaryTree()
a=BinaryTree() # null binary tree

首先制作叶节点，40和60

# if a tree has only root node, it is still binary tree
x.make_tree(40,a,a)
y.make_tree(60,a,a)

创建内部节点

z.make_tree(20,x,a) #left internal
r.make_tree(50,a,y) #right internal
s.make_tree(30,r,a)
t.make_tree(10,z,s)

Answer 19

Python 中的二叉树

 class Tree(object):
    def __init__(self):
        self.data=None
        self.left=None
        self.right=None
    def insert(self, x, root):
        if root==None:
            t=node(x)
            t.data=x
            t.right=None
            t.left=None
            root=t
            return root
        elif x<root.data:
            root.left=self.insert(x, root.left)
        else:
            root.right=self.insert(x, root.right)
        return root

    def printTree(self, t):
        if t==None:
            return

        self.printTree(t.left)
        print t.data
        self.printTree(t.right)

class node(object):
    def __init__(self, x):
        self.x=x

bt=Tree()
root=None
n=int(raw_input())
a=[]
for i in range(n):
    a.append(int(raw_input()))
for i in range(n):
    root=bt.insert(a[i], root)
bt.printTree(root)

Answer 20

class Node:
    """
    single Node for tree
    """

    def __init__(self, data):
        self.data = data
        self.right = None
        self.left = None


class binaryTree:
    """
    binary tree implementation
    """

    def __init__(self):
        self.root = None

    def push(self, element, node=None):
        if node is None:
            node = self.root

        if self.root is None:
            self.root = Node(element)

        else:
            if element < node.data:
                if node.left is not None:
                    self.push(element, node.left)
                else:
                    node.left = Node(element)
            else:
                if node.right is not None:
                    self.push(element, node.right)
                else:
                    node.right = Node(element)

    def __str__(self):
        self.printInorder(self.root)
        return "\n"

    def printInorder(self, node):
        """
        print tree in inorder
        """
        if node is not None:
            self.printInorder(node.left)
            print(node.data)
            self.printInorder(node.right)


def main():
    """
    Main code and logic comes here
    """
    tree = binaryTree()
    tree.push(5)
    tree.push(3)
    tree.push(1)
    tree.push(3)
    tree.push(0)
    tree.push(2)
    tree.push(9)
    tree.push(10)
    print(tree)


if __name__ == "__main__":
    main()