# 110.Balanced Binary Tree

**110.Balanced Binary Tree**

难度:Easy

> 给定一个二叉树，判断它是否是高度平衡的二叉树。

本题中，一棵高度平衡二叉树定义为：

一个二叉树每个节点 的左右两个子树的高度差的绝对值不超过1。

```
示例 1:

给定二叉树 [3,9,20,null,null,15,7]

    3
   / \
  9  20
    /  \
   15   7
返回 true 。

示例 2:

给定二叉树 [1,2,2,3,3,null,null,4,4]

       1
      / \
     2   2
    / \
   3   3
  / \
 4   4
返回 false 。
```

通过树的高度判断。

```
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
private:
    int deep(TreeNode* root)
    {
        if(!root) return 0;
        int r=1,l=1;
        if(!root->left && !root->right) return 1;
        if(root->left) l+=deep(root->left);
        if(root->right) r+=deep(root->right);
       
        if(l<0 || r<0 || abs(l-r) > 1) return INT_MIN;
        return max(l,r);
    }
public:
    bool isBalanced(TreeNode* root) {
        if(!root) return true;
   //     cout<<deep(root)<<endl;
        return deep(root)>0;
        
    }
};
```

> 执行用时 :32ms, 在所有 C++ 提交中击败了39.18%的用户\
> 内存消耗 :17.4MB, 在所有 C++ 提交中击败了21.21%的用户


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