Balanced Binary Tree——数是否是平衡，即任意节点左右字数高度差不超过1

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.

 1 /**
 2  * Definition for binary tree
 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     bool isBalanced(TreeNode *root) {
13         if(root==NULL) return true;
14         int l=height(root->left);
15         int r=height(root->right);
16         if(l<0||r<0||l-r>1||r-l>1) return false;
17         return true;
18     }
19     int height(TreeNode *root){
20         if(root==NULL) return 0;
21         int l=height(root->left);
22         int r=height(root->right);
23         if(l<0||r<0||l-r>1||r-l>1) return -1;
24         return max(l,r)+1;
25     }
26 };

其他做法：

转自：http://blog.csdn.net/feliciafay/article/details/18348065

 1 // 104ms过大集合  
 2 /** 
 3  * Definition for binary tree 
 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 height(TreeNode* root) {  
14         if(root==NULL)  
15             return 0;  
16         else{  
17             int l=height(root->left);  
18             int r=height(root->right);  
19             return 1+((l>r)?l:r);  
20         }  
21     }  
22     bool isBalanced(TreeNode *root) {  
23         if(root==NULL)  
24             return true;  
25         else{  
26             int l,r;  
27             l=height(root->left);  
28             r=height(root->right);  
29             if((l>r+1)||(r>l+1))  
30                 return false;  
31             else  
32                 return isBalanced(root->left)&&isBalanced(root->right);  
33         }  
34     }  
35 };

 1 // 68ms过大集合  简洁版
 2 public:  
 3 int cntHeight(TreeNode *root) {  
 4     if(root == NULL) return 0;  
 5     int l = cntHeight(root->left);  
 6     int r = cntHeight(root->right);  
 7     if(l < 0 || r < 0 || abs(l-r) > 1) return -1; //自定义 return -1，表示不平衡的情况  
 8     else  return max(l, r) + 1;  
 9 }  
10 bool isBalanced(TreeNode *root) {  
11     if(root == NULL) return true;      
12     int l = cntHeight(root->left);  
13     int r = cntHeight(root->right);  
14     if(l < 0 || r < 0 || abs(l-r) > 1) return false;  
15     else return true;  
16 }

 1 // 68ms过大集合  更简洁版
 2 public:  
 3 int cntHeight(TreeNode *root) {  
 4     if(root == NULL) return 0;  
 5     int l = cntHeight(root->left);  
 6     int r = cntHeight(root->right);  
 7     if(l < 0 || r < 0 || abs(l-r) > 1) return -1; //自定义 return -1，表示不平衡的情况  
 8     else  return max(l, r) + 1;  
 9 }  
10 bool isBalanced(TreeNode *root) {  
11     if(root == NULL) return true;      
12     int l = cntHeight(root->left);  
13     int r = cntHeight(root->right);  
14     if(l < 0 || r < 0 || abs(l-r) > 1) return false;  
15     else return true;  
16 }