leetcode 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.

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不明白这个题为什么自己还是想了很久。

改了很多，有的时候思路不明白还是写下来可以帮助自己理清头绪。

就是求左子树的高度，右子树的高度，假如左子树减右子树的高度的绝对值大于2就不是平衡二叉树了，如果顶点满足要求，就要看左子树是否满足要求，然后右子树。

不知道为什么写错了函数，写了一个专门求左子树高度的函数，专门求右子树的函数，脑子一定是被门夹了。

 1 #include<math.h>
 2 /**
 3  * Definition for a binary tree node.
 4  * struct TreeNode {
 5  *     int val;
 6  *     TreeNode *left;
 7  *     TreeNode *right;
 8  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 9  * };
10  */
11 class Solution {
12 public:
13     int maxDepth(TreeNode* root) {
14         int lengthleft=0,lengthright=0;
15         if (root==NULL) return 0;
16         if (root->left == NULL && root->right == NULL){
17             return 1;
18         }
19 
20         if(root->left!=NULL){ 
21             lengthleft=maxDepth(root->left)+1;
22 
23         }
24         if(root->right!=NULL){
25             lengthright=maxDepth(root->right)+1;
26         }
27             
28         return lengthright>lengthleft?lengthright:lengthleft;
29             
30         }
31     
32     bool isBalanced(TreeNode* root) {
33         if(root==NULL) return true;
34         int m=maxDepth(root->left)-maxDepth(root->right);
35         cout<<m<<endl;
36         if(abs(m)<2) return isBalanced(root->left)&&isBalanced(root->right);
37         else return false;
38     }
39     
40 };