使用父指针删除二叉搜索树

Question

我已经尝试解决在线挑战 9 天了。我有一个重复 100,000 次的在线插入删除，我需要找到其中的中位数。我尝试了两个堆，但实现随机删除不起作用。我现在开始使用二叉搜索树，因为它们似乎在 2 秒内在我的计算机上插入和删除了 100,000 个数据。我正在研究 python，我需要在 16 秒内运行。

我已经在线检查了解决方案，但完全不能满足我的需要。我发现在插入和删除时计算中位数可能是一个好策略。这我写了两个方法来获取节点的顺序后继或前驱。

问题是我认为是 - 在递归删除期间不正确地分配父指针。

我尝试了两种技术，但都效果不佳，如果我能得到一些帮助，我将不胜感激！

代码：

import sys
class BSTNode:
    def __init__(self,x,parent):
        self.data = x
        self.left = None
        self.right = None
        self.count = 1
        self.parent = parent

def delete(x,T):
    if T is None:
        print('Element Not Found')
    elif x<T.data:
        T.left = delete(x,T.left)
    elif x>T.data:
        T.right = delete(x,T.right)
    elif T.left and T.right:
        TempNode = findMin(T.right)
        T.data = TempNode.data
        T.right = delete(TempNode.data,T.right)
    else:
        if T.left is None:
            T = T.right
        elif T.right is None:
            T = T.left
    return T

def findMin(T):
    if T.left:
        return findMin(T.left)
    else:
        return T

def insert(x,T,parent=None):
    if T is None:
        T = BSTNode(x,parent)
    elif x<T.data:
        T.left = insert(x,T.left,T)
    elif x>T.data:
        T.right = insert(x,T.right,T)
    else:
        T.count = T.count + 1
    return T

def inorder(T):
    if T is None:
        return
    else:
        inorder(T.left)
        b = back(T)
        if b:
            print("back:",b.data)
        print(T.data)
        n = next(T)
        if n:
            print("next:",n.data)
        inorder(T.right)

def preorder(T,i=0):
    if T is None:
        return
    else:
        for j in range(i):
            sys.stdout.write("    ")
        print(T.data)
        preorder(T.left,i+1)
        preorder(T.right,i+1)

def next(node):
    if node is None:
        return
    if node.right:
        n = node.right
        while n.left:
            n = n.left
        return n
    else:
        n = node
        while n.parent and n.parent.right is n:
            n = n.parent
        if n.parent and n.parent.left is n:
            return n.parent
        else:
            return

def back(node):
    if node is None:
        return
    if node.left:
        n = node.left
        while n.right:
            n = n.right
        return n
    else:
        n = node
        while n.parent and n.parent.left is n:
            n = n.parent
        if n.parent and n.parent.right is n:
            return n.parent
        else:
            return

T = None
T = insert(7,T)
T = insert(4,T)
T = insert(2,T)
T = insert(1,T)

T = insert(13,T)
T = insert(15,T)
T = insert(16,T)
T = insert(6,T)

T = insert(5,T)
T = insert(3,T)
T = insert(11,T)
T = insert(14,T)

T = insert(12,T)
T = insert(9,T)
T = insert(8,T)
T = insert(10,T)

T = delete(11,T)
T = delete(12,T)
T = delete(13,T)
T = delete(8,T)
preorder(T)
inorder(T)

输出

7
    4
        2
            1
            3
        6
            5
    14
        9
            10
        15
            16
1
('next:', 2)
('back:', 1)
2
('next:', 3)
('back:', 2)
3
('next:', 4)
('back:', 3)
4
('next:', 5)
('back:', 4)
5
('next:', 6)
('back:', 5)
6
('next:', 7)
('back:', 6)
7
('next:', 9)
9
('next:', 10)
('back:', 9)
10
('next:', 12)
('back:', 10)
14
('next:', 15)
('back:', 14)
15
('next:', 16)
('back:', 15)
16

预期 - 9 的后面是 7

我的回答

def delete(x,T,parent=None):
    if T is None:
        print('Element Not Found')
    elif x<T.data:
        T.left = delete(x,T.left,T)
    elif x>T.data:
        T.right = delete(x,T.right,T)
    elif T.count==1:
        # 2 CHILDREN
        if T.left and T.right:
            TempNode = findMin(T.right)
            T.data = TempNode.data
            T.right = delete(TempNode.data,T.right,T)
        # 0 CHILDREN
        elif T.left is None and T.right is None:
            T = None
        # 1 CHILDREN
        elif T.right is not None:
            T = T.right
            T.parent = parent
        elif T.left is not None:
            T = T.left
            T.parent = parent
    else:
        T.count = T.count - 1
    return T

现在

9
    4
        2
            1
            3
        5
    14
        10
        15
            16
1 (2) 
('next:', 2)
('back:', 1)
2 (4) 
('next:', 3)
('back:', 2)
3 (2) 
('next:', 4)
('back:', 3)
4 (9) 
('next:', 5)
('back:', 4)
5 (4) 
('next:', 9)
('back:', 5)
9  
('next:', 10)
('back:', 9)
10 (14) 
('next:', 14)
('back:', 10)
14 (9) 
('next:', 15)
('back:', 14)
15 (14) 
('next:', 16)
('back:', 15)
16 (15)

Answer 1

您似乎没有在delete(x,T): 例程的else: 部分中的T = T.right 和T = T.left 之后更新T 的父指针。另一种选择可能是在替换左子或右子时更新父指针。我认为在整个代码中始终遵循这两种约定中的任何一种都可以解决您的问题。