A)  3            B)    3 
   / \                  \
  9  20                 20
    /  \                / \
   15   7              15  7

解题思路：

    由于需要每个子树的高度信息，所以需要增加一个高度函数，用来返回当前节点的高度。然后再用高度差信息判断当前节点是否为平衡节点，是则继续向下判断，否则返回false.

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */

class Solution {
public:
    /**
     * @param root: The root of binary tree.
     * @return: True if this Binary tree is Balanced, or false.
     */
    bool isBalanced(TreeNode * root)
    {
        // write your code here
        if(root == NULL)
            return true;
        
        if(abs(hight(root->left)-hight(root->right))>1)//当前节点的两个子树的高度差超过1了
            return false;
        else
            return isBalanced(root->left) && isBalanced(root->right);   
    }
    
    int hight(TreeNode * root)
    {
        if(root == NULL)
            return 0;
        
        int res = max(hight(root->left) , hight(root->right));
        return res+1;
    }
};