leetcode validBST

算法:通过递归并限定上下限元素的值的范围可以计算


package com.bupt.acm.leetcode;

public class ValidBinary {

private class TreeNode{
int val;
TreeNode left;
TreeNode right;
TreeNode(int x){
val=x;
}
}
public boolean isValidBST(TreeNode root){
if(root==null){
return true;
}
return validBST(root,Integer.MIN_VALUE,Integer.MAX_VALUE);
}
private boolean validBST(TreeNode root, int min, int max) {
// TODO Auto-generated method stub
if(root==null){
return true;
}
if(root.val>min&&root.val<max){
return validBST(root.left,min,root.val)&&validBST(root.right, root.val, max);
}else{
return false;
}
}

}

leetcode validBST,布布扣,bubuko.com

时间: 2024-10-11 22:07:21

leetcode validBST的相关文章

[LeetCode]题解(python):098 Validate Binary Search Tree

题目来源 https://leetcode.com/problems/validate-binary-search-tree/ Given a binary tree, determine if it is a valid binary search tree (BST). Assume a BST is defined as follows: The left subtree of a node contains only nodes with keys less than the node'

[leetcode]Validate Binary Search Tree @ Python

原题地址:https://oj.leetcode.com/problems/validate-binary-search-tree/ 题意:检测一颗二叉树是否是二叉查找树. 解题思路:看到二叉树我们首先想到需要进行递归来解决问题.这道题递归的比较巧妙.让我们来看下面一棵树: 4 /    \ 2 6 /    \   /   \  1      3 5    7 对于这棵树而言,怎样进行递归呢?root.left这棵树的所有节点值都小于root,root.right这棵树的所有节点值都大于roo

[LeetCode] 349 Intersection of Two Arrays &amp; 350 Intersection of Two Arrays II

这两道题都是求两个数组之间的重复元素,因此把它们放在一起. 原题地址: 349 Intersection of Two Arrays :https://leetcode.com/problems/intersection-of-two-arrays/description/ 350 Intersection of Two Arrays II:https://leetcode.com/problems/intersection-of-two-arrays-ii/description/ 题目&解法

LeetCode 442. Find All Duplicates in an Array (在数组中找到所有的重复项)

Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all the elements that appear twice in this array. Could you do it without extra space and in O(n) runtime? Example: Input: [4,3,2,7,

LeetCode OJ - Sum Root to Leaf Numbers

这道题也很简单,只要把二叉树按照宽度优先的策略遍历一遍,就可以解决问题,采用递归方法越是简单. 下面是AC代码: 1 /** 2 * Sum Root to Leaf Numbers 3 * 采用递归的方法,宽度遍历 4 */ 5 int result=0; 6 public int sumNumbers(TreeNode root){ 7 8 bFSearch(root,0); 9 return result; 10 } 11 private void bFSearch(TreeNode ro

LeetCode OJ - Longest Consecutive Sequence

这道题中要求时间复杂度为O(n),首先我们可以知道的是,如果先对数组排序再计算其最长连续序列的时间复杂度是O(nlogn),所以不能用排序的方法.我一开始想是不是应该用动态规划来解,发现其并不符合动态规划的特征.最后采用类似于LRU_Cache中出现的数据结构(集快速查询和顺序遍历两大优点于一身)来解决问题.具体来说其数据结构是HashMap<Integer,LNode>,key是数组中的元素,所有连续的元素可以通过LNode的next指针相连起来. 总体思路是,顺序遍历输入的数组元素,对每个

LeetCode OJ - Surrounded Regions

我觉得这道题和传统的用动规或者贪心等算法的题目不同.按照题目的意思,就是将被'X'围绕的'O'区域找出来,然后覆盖成'X'. 那问题就变成两个子问题: 1. 找到'O'区域,可能有多个区域,每个区域'O'都是相连的: 2. 判断'O'区域是否是被'X'包围. 我采用树的宽度遍历的方法,找到每一个'O'区域,并为每个区域设置一个value值,为0或者1,1表示是被'X'包围,0则表示不是.是否被'X'包围就是看'O'区域的边界是否是在2D数组的边界上. 下面是具体的AC代码: class Boar

LeetCode 10. Regular Expression Matching

https://leetcode.com/problems/regular-expression-matching/description/ Implement regular expression matching with support for '.' and '*'. '.' Matches any single character. '*' Matches zero or more of the preceding element. The matching should cover

(leetcode题解)Pascal&#39;s Triangle

Pascal's Triangle  Given numRows, generate the first numRows of Pascal's triangle. For example, given numRows = 5,Return [ [1], [1,1], [1,2,1], [1,3,3,1], [1,4,6,4,1] ] 题意实现一个杨辉三角. 这道题只要注意了边界条件应该很好实现出来,C++实现如下 vector<vector<int>> generate(int