Leetcode 110 Balanced Binary Tree 二叉树

判断一棵树是否是平衡树，即左右子树的深度相差不超过1.

我们可以回顾下depth函数其实是Leetcode 104 Maximum Depth of Binary Tree 二叉树

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     int depth(TreeNode* root){
13         if(!root) return 0;
14         else{
15             return max(depth(root->left), depth(root->right)) + 1;
16         }
17     }
18     bool isBalanced(TreeNode* root) {
19         if(!root) return true;
20         else if(abs(depth(root->left) - depth(root->right))>1 ){
21             return false;
22         }
23         else return isBalanced(root->left) && isBalanced(root->right);
24     }
25 };

 

posted @ 2016-03-01 19:10  Breeze0806  阅读(130)  评论(0)  编辑  收藏  举报