LeetCode学习篇二十五——Balanced Binary Tree

题目：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.
通过率：35.9% 难度：easy
这道题就是要判断所给的二叉树是不是平衡二叉树。平衡二叉树（Balanced Binary Tree）又被称为AVL树（有别于AVL算法），且具有以下性质：它是一 棵空树或它的左右两个子树的高度差的绝对值不超过1，并且左右两个子树都是一棵平衡二叉树。所以实现方法通过两层递归，一个递归是求树的高度，另一个递归是判断左右子树是否高度相差不超过1，代码实现如下：

/**
 * 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 {
public:
    int height(TreeNode *root) {
        if(root == NULL) {
            return 0;
        }
        else return max(height(root->left),height(root->right))+1;
    }
    bool isBalanced(TreeNode* root) {
        if(root == NULL) {
            return true;
        }
        else if(height(root->left)-height(root->right) <= 1 && height(root->left)-height(root->right) >= -1)  {
            return isBalanced(root->left) && isBalanced(root->right);
        }
        else return false;
    }
};