刷题-- 平衡二叉树

2019年3月16日 270次阅读 来源: 平衡二叉树

输入一棵二叉树，判断该二叉树是否是平衡二叉树。

基本思想：递归。

# -*- coding:utf-8 -*-
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
    def Depth(self, pRoot):
        if pRoot == None:
            return 0
        else:
            left = self.Depth(pRoot.left)
            right = self.Depth(pRoot.right)
            return max(left, right) + 1


    def IsBalanced_Solution(self, pRoot):
        # write code here
        if pRoot == None:
            return True
        else:
            if self.IsBalanced_Solution(pRoot.left) and self.IsBalanced_Solution(pRoot.right):
                if abs(self.Depth(pRoot.left) - self.Depth(pRoot.right)) <= 1:
                    return True
                else:
                    return False

上面这种做法会使得有些结点被重复遍历，效率不高。
改进版：每个结点只遍历一遍。

# -*- coding:utf-8 -*-
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
    def IsBalanced(self, pRoot):
        # write code here
        if pRoot == None:
            return True, 0
        else:
            res_left = self.IsBalanced(pRoot.left)
            res_right = self.IsBalanced(pRoot.right)

            if res_left[0] and res_right[0]:
                if abs(res_left[1] - res_right[1]) <= 1:
                    return True, max(res_left[1], res_right[1]) + 1

            return False, 0


    def IsBalanced_Solution(self, pRoot):
        # write code here
        return self.IsBalanced(pRoot)[0]
    原文作者：平衡二叉树
    原文地址: https://blog.csdn.net/Treasure_Z/article/details/80462498
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