Leetcode 110. Balanced Binary Tree

110. Balanced Binary Tree

  • Total Accepted: 119517
  • Total Submissions:345415
  • Difficulty: Easy

 

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.

 

思路：
方法一：直接按照定义做。平衡树的定义：左右子树高差的绝对值不大于1且左右子树都是平衡树。

方法二：设计了isDepth函数：如果root为根的树是平衡树，则返回树高；否则返回-1。

 

 

代码：
方法一：

 1 class Solution {
 2 public:
 3     int Depth(TreeNode* root){
 4         if(!root){
 5             return 0;
 6         }
 7         return 1+max(Depth(root->left),Depth(root->right));
 8     }
 9     bool isBalanced(TreeNode* root) {
10         if(!root){
11             return true;
12         }
13         int left=Depth(root->left);
14         int right=Depth(root->right);
15         return abs(left-right)<=1&&isBalanced(root->left)&&isBalanced(root->right);    
16     }
17 };

 

方法二：

 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     //如果root为根的树是平衡树，则返回树高；否则返回-1
13     int isDepth(TreeNode* root){
14         if(!root){
15             return 0;
16         }
17         int left=isDepth(root->left);
18         if(left==-1) return -1;
19         int right=isDepth(root->right);
20         if(right==-1) return -1;
21         if(abs(left-right)<=1){
22             return max(left,right)+1;
23         }
24         return -1;
25     }
26     bool isBalanced(TreeNode* root) {
27         return isDepth(root)!=-1;
28     }
29 };

 

posted @ 2016-07-09 14:31  Deribs4  阅读(235)  评论(0)  编辑  收藏  举报