# 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.

题目翻译:
给定一颗二叉树, 写一个函数来检测这棵树是否是平衡二叉树. 对于这个问题, 一颗平衡树的定义是其中任意节点的左右子树的高度差不大于1.

解题思路:
这道题其实就是应用DFS,对于一颗二叉树边计算树的高度边计算差值,针对树里面的每一个节点计算它的左右子树的高度差,如果差值大于1,那么就返回-1,如果不大于1,从下往上再次检测.

时间复杂度:
由于是运用DFS,所以时间复杂度为O(n).

代码如下:
```c++

/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool isBalanced(TreeNode *root) {
//corner case check
if(root == NULL)
return true;
int isBalanced = getHeight(root);
if(isBalanced != -1)
return true;
else
return false;
}

int getHeight(TreeNode* root)
{
if(root == NULL)
return 0;
int leftHeight = getHeight(root->left);
if(leftHeight == -1)
return -1;
int rightHeight = getHeight(root->right);
if(rightHeight == -1)
return -1;
int diffHeight = rightHeight > leftHeight? rightHeight-leftHeight:leftHeight-rightHeight;
if(diffHeight > 1)
return -1;
else
return diffHeight = (rightHeight>leftHeight?rightHeight:leftHeight)+1;
}
};
```