Balanced Binary Tree（平衡二叉树）

来源：https://leetcode.com/problems/balanced-binary-tree

Given a binary tree, determine if it is height-balanced.

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

平衡二叉树：是它一棵空树或它的左右两个子树的高度差的绝对值不超过1，并且左右两个子树都是一棵平衡二叉树。

Java

 /**
  * Definition for a binary tree node.
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 class Solution {
     private boolean isBalancedFlag = true;
     private int getDepth(TreeNode root) {
         if(root == null) {
             return 0;
         }
         int leftDepth = getDepth(root.left) + 1;
         int rightDepth = getDepth(root.right) + 1;
         if(Math.abs(leftDepth - rightDepth) > 1) {
             isBalancedFlag = false;
         }
         return leftDepth > rightDepth ? leftDepth : rightDepth;
     }
     public boolean isBalanced(TreeNode root) {
         getDepth(root);
         return isBalancedFlag;
     }
 }// 1 ms

Python

 # -*- coding:utf-8 -*-
 # class TreeNode:
 #     def __init__(self, x):
 #         self.val = x
 #         self.left = None
 #         self.right = None
 class Solution:
     __is_balanced = True
     def getDepth(self, pRoot):
         if pRoot == None:
             return 0
         left_depth = self.getDepth(pRoot.left)
         right_depth = self.getDepth(pRoot.right)
         if abs(left_depth-right_depth) > 1:
             self.__is_balanced = False
         return left_depth+1 if left_depth>right_depth else right_depth+1
     def IsBalanced_Solution(self, pRoot):
         self.getDepth(pRoot)
         return self.__is_balanced