【LeetCode】110. Balanced Binary Tree-判断是否为平衡二叉树

 1 /**
 2  * Definition for a binary tree node.
 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){
13             return true;
14         }
15         int left = depth(root.left);
16         int right = depth(root.right);
17         if(left+1<right || right+1<left){
18             return false;
19         }else{
20             return isBalanced(root.left)&&isBalanced(root.right);   
21         }
22     }
23     
24     public int depth(TreeNode root){
25         if(root==null){
26             return 0;
27         }else{
28             int left = depth(root.left);
29             int right = depth(root.right);
30             return left<right?right+1:left+1;
31         }
32     }
33 }