110leetcode 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.

链接：https://leetcode.com/problems/balanced-binary-tree/description/

题意很简单，判断一棵树是否为平衡二叉树，左右高度是否相差超过1

思路：

从root开始，计算每个节点的左右子树的高度，判断相差是否超过1，只要简单的两个递归就可以了，细节看代码

ac代码：

/**
 * 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:
    bool isBalanced(TreeNode* root) {
        if(root==NULL)    //判断是否为空，为空则返回true
            return true;
        if(abs(Height(root->left)-Height(root->right))>1)  //比较左右子树的高度是否相差超过1,
            return false;
        return isBalanced(root->left)&&isBalanced(root->right);  //遍历整棵树的左右子树节点，判断每个分支是否为平衡二叉树
    }
    int Height(TreeNode* root){
        if(root==NULL)
            return 0;  //若为空，则返回高度为0；
        int left=1+Height(root->left);  //左子树加上节点自己
        int right=1+Height(root->right);  //同理
        return max(left,right);   //返回左右子树最大的那个
    }
};