39平衡二叉树判断python

2019年3月16日 298次阅读 来源: 平衡二叉树

题目：输入一棵二叉树，判断该二叉树是否是平衡二叉树。若左右子树深度差不超过1则为一颗平衡二叉树。
思路：1、使用获取二叉树深度的方法来获取左右子树的深度
2、左右深度相减，若大于1返回False
3、通过递归对每个节点进行判断，若全部均未返回False，则返回True

# -*- coding:utf-8 -*-
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
    def getDeepth(self, Root):
        if Root is None:
            return 0
        nright = self.getDeepth(Root.right)
        nleft = self.getDeepth(Root.left)
        return max(nright, nleft)+1
    def IsBalanced_Solution(self, pRoot):
        # write code here
        if pRoot is None:
            return True
        right = self.getDeepth(pRoot.right)
        left = self.getDeepth(pRoot.left)
        if abs(right - left) >1:
            return False
        return self.IsBalanced_Solution(pRoot.right) and self.IsBalanced_Solution(pRoot.left)

进阶思路：
采用后续遍历，当遍历到根节点时，每个节点只会遍历一次
以下程序无法运行，仅提供思路

class Solution:
    def IsBalanced(self, pRoot, depth):
        if pRoot is None:
            return True
        if self.IsBalanced(pRoot.right, right) and self.IsBalanced(pRoot.left, left):
            diff = abs(left - right)
            if diff <= 1:
                depth = max(left, right) +1
                return True
        return False
    def IsBalanced_Solution(self, pRoot):
        # write code here
        depth = 0
        return self.IsBalanced(pRoot, depth)
    原文作者：平衡二叉树
    原文地址: https://blog.csdn.net/slibra_L/article/details/78173760
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