C++版 - 剑指offer 面试题39：判断平衡二叉树(LeetCode 110. Balanced Binary Tree) 题解

剑指offer 面试题39：判断平衡二叉树

提交网址：  http://www.nowcoder.com/practice/8b3b95850edb4115918ecebdf1b4d222?tpId=13&tqId=11192

时间限制：1秒       空间限制：32768K      参与人数：2481

题目描述

输入一棵二叉树，判断该二叉树是否是平衡二叉树。

分析：

平衡二叉树定义

递归解法 AC代码：

#include<iostream>
#include<vector>
using namespace std;
struct TreeNode{
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution {
public:
	int getHeight(TreeNode *root)
	{
		if(root==NULL) return 0;
		int lh=getHeight(root->left);
		int rh=getHeight(root->right);
		int height=(lh<rh)?(rh+1):(lh+1);  // 记住：要加上根节点那一层，高度+1
		return height;
	}
    bool IsBalanced_Solution(TreeNode* pRoot) {
		if(pRoot==NULL) return true;
		int lh=getHeight(pRoot->left);
		int rh=getHeight(pRoot->right);
		if(lh-rh>1 || lh-rh<-1)	return false;
		else return (IsBalanced_Solution(pRoot->left) && IsBalanced_Solution(pRoot->right)); // 继续递归判断
    }
};
// 以下为测试
int main()
{
	Solution sol;
    TreeNode *root = new TreeNode(1);
    root->right = new TreeNode(2);
    root->right->left = new TreeNode(3);
    bool res=sol.IsBalanced_Solution(root);
    cout<<res<<" ";
    return 0;
}

LeetCode 110. Balanced Binary Tree

Total Accepted: 111269 Total Submissions: 326144 Difficulty: Easy

提交网址: https://leetcode.com/problems/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.

AC代码：

class Solution {
public:
	int getHeight(TreeNode *root)
	{
		if(root==NULL) return 0;
		int lh=getHeight(root->left);
		int rh=getHeight(root->right);
		int height=(lh<rh)?(rh+1):(lh+1);  // 记住：要加上根节点那一层，高度+1
		return height;
	}
    bool isBalanced(TreeNode *root) {
		if(root==NULL) return true;
		int lh=getHeight(root->left);
		int rh=getHeight(root->right);
		if(lh-rh>1 || lh-rh<-1)	return false;
		else return (isBalanced(root->left) && isBalanced(root->right)); // 继续递归判断
    }
};