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:

  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.     bool isBalanced(TreeNode *root) {
  14.         //corner case check
  15.         if(root == NULL)
  16.             return true;
  17.         int isBalanced = getHeight(root);
  18.         if(isBalanced != -1)
  19.             return true;
  20.         else
  21.             return false;
  22.     }
  24.     int getHeight(TreeNode* root)
  25.     {
  26.         if(root == NULL)
  27.             return 0;
  28.         int leftHeight = getHeight(root->left);
  29.         if(leftHeight == -1)
  30.             return -1;
  31.         int rightHeight = getHeight(root->right);
  32.         if(rightHeight == -1)
  33.             return -1;
  34.         int diffHeight = rightHeight > leftHeight? rightHeight-leftHeight:leftHeight-rightHeight;
  35.         if(diffHeight > 1)
  36.             return -1;
  37.         else
  38.             return diffHeight = (rightHeight>leftHeight?rightHeight:leftHeight)+1;
  39.     }
  40. };