| 标题： |
Balanced Binary Tree |
| 通过率： | 32.3% |
| 难度： | 简单 |
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。这个可以由二叉树深度来计算，再二叉树深度计算中每次进行深度比较取深度较深的一个，那么平衡二叉树再比较深度时就是A-B绝对值超过一的话就不是二叉树了。立马返回-1 ，当左右子树有一边深度值是-1时立马终止，返回此树非平衡二叉树，如果A-B绝对值总是在0，1，-1三个数之中，那么继续深度递归判断，具体代码如下：
1 /** 2 * Definition for binary tree 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 public class Solution { 11 public boolean isBalanced(TreeNode root) { 12 if(root==null) return true; 13 if(height(root)==-1) return false; 14 else return true; 15 16 } 17 public int height(TreeNode root){ 18 if(root==null) return 0; 19 int depth1=0; 20 int depth2=0; 21 depth1=height(root.right); 22 depth2=height(root.left); 23 if(depth1==-1||depth2==-1) return -1; 24 if((depth1-depth2>1 )||(depth1-depth2)<-1) 25 return -1; 26 else{ 27 if(depth1>depth2) 28 return depth1+1; 29 else 30 return depth2+1; 31 } 32 33 } 34 }