二叉树总结及部分Lintcode题目分析 1

1. 遍历问题 Preorder / Inorder / Postorder

preorder: root left right

inorder: left root right

postorder: left right root

遇到二叉树的问题，就是一步步的拆分，最常用的就是Divide & Conquer的递归实现

贴一个preorder的例子

Definition of TreeNode:
class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left, self.right = None, None
"""


class Solution:
    """
    @param root: The root of binary tree.
    @return: Preorder in ArrayList which contains node values.
    """
    def preorderTraversal(self, root):
        result = []
        if root is None:
            return result
        result = [root.val] + self.preorderTraversal(root.left) + self.preorderTraversal(root.right)
        return result

 

2. 遍历问题的变形，比如Maximum Depth of Binary Tree，是同样的思想，从root出发的最深路径一定是左右子数最深路径的最大值。就可以一步步把树分拆下去。

 

贴一个代码

 

"""
Definition of TreeNode:
class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left, self.right = None, None
"""
class Solution:
    """
    @param root: The root of binary tree.
    @return: An integer
    """ 
    def maxDepth(self, root):
        depth = 0
        if root is None:
            return depth
        depth = max(self.maxDepth(root.left), self.maxDepth(root.right)) + 1
        return depth
            

 

 3. 继续进行延伸，在Balanced binary tree 的做法中可以应用 maxDepth的解法

判断是否是balanced binary tree最直观的就是左右子树的高度差，对任意一个点来说的话 left - right > 1 则不是balanced

"""
Definition of TreeNode:
class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left, self.right = None, None
"""
class Solution:
    """
    @param root: The root of binary tree.
    @return: True if this Binary tree is Balanced, or false.
    """
    def maxDepth(self, root):
        if root is None:
            return 0
        left = self.maxDepth(root.left)
        right = self.maxDepth(root.right)
        # at first, the abs is ignored and the operations when the maxdepth is calculated is wrong
        if left == -1 or right == -1 or abs(left - right) > 1:
            return -1
        return max(left, right) + 1
        
    def isBalanced(self, root):
        if root is None:
            return True
        return self.maxDepth(root) != -1