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
}
};