572. Subtree of Another Tree(easy)】的更多相关文章

Given two non-empty binary trees s and t, check whether tree t has exactly the same structure and node values with a subtree of s. A subtree of s is a tree consists of a node in s and all of this node's descendants. The tree s could also be considere…
Given two non-empty binary trees s and t, check whether tree t has exactly the same structure and node values with a subtree of s. A subtree of s is a tree consists of a node in s and all of this node's descendants. The tree s could also be considere…
Given two binary trees, write a function to check if they are equal or not. Two binary trees are considered equal if they are structurally identical and the nodes have the same value. 思路:太简单! bool isSameTree(TreeNode *p, TreeNode *q) { if(p == NULL &…
转自:http://www.wowotech.net/linux_kenrel/dt-code-analysis.html Device Tree(三):代码分析 作者:linuxer 发布于:2014-6-6 16:03 分类:统一设备模型 一.前言 Device Tree总共有三篇,分别是: 1.为何要引入Device Tree,这个机制是用来解决什么问题的?(请参考引入Device Tree的原因) 2.Device Tree的基础概念(请参考DT基础概念) 3.ARM linux中和De…
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"> <html> <head> <title>Tree(树)</title> <meta http-equiv="content-type" content="text/html; charset=UTF-8"> <link rel="s…
EASY(easy) sol:非常经典的题,取了一次之后,把线段树上这一段变成相反数 然后再贪心取和最大的. 重复以上操作,发现最后一定有对应的解,且根据贪心过程一定 是最大的 线段树上维护区间和最大/小及位置,左/右连续最大/小及位置, 取反标记 除了写起来特别麻烦之外都还好 #include <bits/stdc++.h> using namespace std; typedef int ll; inline ll read() { ll s=; ; char ch=' '; while(…
2019.7.11leetcode刷题 难度 easy 题目名称 回文数 题目摘要 判断一个整数是否是回文数.回文数是指正序(从左向右)和倒序(从右向左)读都是一样的整数. 思路 一些一定不为回文数的数:1.负数2.大于0,但末位为0的数(x> 0 && x % 10 == 0)如果是上面这些情况则直接返回false. 将整数的每一位存入数组中,arr[i]代表整数的倒数i+1位. 然后判断arr[i]是否和arr[num -1 -i]位是否相等,如果相等则判断下一位:否则返回fal…
141. Linked List Cycle(Easy)2019.7.10 题目地址https://leetcode.com/problems/linked-list-cycle/ Given a linked list, determine if it has a cycle in it. To represent a cycle in the given linked list, we use an integer pos which represents the position (0-i…
2021.07.19 BZOJ2654 tree(生成树) tree - 黑暗爆炸 2654 - Virtual Judge (vjudge.net) 重点: 1.生成树的本质 2.二分 题意: 有一张无相带权连通图,每一条边都是黑色或者白色,求一棵恰好有need条白色边的最小生成树. 分析: 我们可以把每一条边按照一定优先级进行排序,当然,手动优先级(手动狗头),对于同一个权值的边,我们规定白色边比黑色边优先级高.但是对于权值特别小的的黑色边,在构成一棵生成树时,搞不好所有边都是由黑色边构成的…
题目链接 描述 Little Valentine liked playing with binary trees very much. Her favorite game was constructing randomly looking binary trees with capital letters in the nodes.  This is an example of one of her creations:  D / \ / \ B E / \ \ / \ \ A C G / /…