Parliament Time limit: 1.0 secondMemory limit: 64 MB A new parliament is elected in the state of MMMM. Each member of the parliament gets his unique positive integer identification number during the parliament registration. The numbers were given in…

二叉树水题,特别是昨天刚做完二叉树用中序后序建树,现在来做这个很快的. 跟昨天那题差不多,BST后序遍历的特型,找到最后那个数就是根,向前找,比它小的那块就是他的左儿子,比它大的那块就是右儿子,然后递归儿子继续建树. 代码: #include <cstdio> #include <cstdlib> const int maxn = 70000; struct Node { int v; Node *l; Node *r; }; int arr[maxn]; bool flag =…

1136 先由后左 再父 建一个二叉树 #include <iostream> #include<cstdio> #include<cstring> #include<stdlib.h> #include<algorithm> using namespace std; #define N 3010 int a[N]; int tr[N]; int g; void build(int s,int te,int ro) { if(s==te) { g…

题目链接:http://acm.timus.ru/problem.aspx?space=1&num=1136 题目描述:给定一个按照(左子树-右子树-根)(即先序)遍历序列的树,求其按照 右子树-左子树-根 遍历的结果.(每个数都不同) 题目思路:按照题目意思其实构造的是一个二叉查找树,满足左子树元素都不大于当前根的元素,右子树元素都不小于当前根的元素. 而且二叉查找树按照 中序遍历 的结果是元素按照递增顺序输出(二叉查找树的性质).所以实际上又告诉了你中序遍历的结果(即把所给元素递增排序的结果…

Parliament Time limit: 1.0 secondMemory limit: 64 MB A new parliament is elected in the state of MMMM. Each member of the parliament gets his unique positive integer identification number during the parliament registration. The numbers were given in…

题意 输入一棵树的后缀表达式(按左-右-中顺序访问),这棵树的每一个结点的数值都比它的左子树结点的数值大,而比它的右子树结点的数值小,要求输出其按右-左-中顺序访问的表达式.所有的数都为正整数,而且不会重复. 思路 很像根据中缀和后缀表达式求前缀表达式之类的题.方法自然也差不多.由DFS的括号性质可知,每一个树都对应表达式的一个区间,而此题中区间的最后一个就是树的根,然后根据根的大小可以把区间分为值小于和大于根值的两部分,即左子树和右子树,然后递归地输出右子树.左子树,最后再输出根即可. 代码…

剑指offer 重建二叉树 提交网址: http://www.nowcoder.com/practice/8a19cbe657394eeaac2f6ea9b0f6fcf6?tpId=13&tqId=11157 或 leetcode 105: https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/ 参与人数:5246  时间限制:1秒  空间限制:32768K 题目描述 输入某…

PAT 1139 1138 1137 1136 一个月不写题,有点生疏..脑子跟不上手速,还可以啦,反正今天很开心. PAT 1139 First Contact 18/30 找个时间再修bug 23/30 28/30 30/30 补完啦 这道题的细节坑点: 1.输出id需要补全4位 用print("%04d") 5分ok 2.需要舍弃查询的相恋男女的直接边 3分ok 3.大坑 0000 与 -0000 使用int型无法区分男女的. 2分ok 想法:用string输入 看长度和第一位符…

二叉树(Binary Tree)是最简单的树形数据结构,然而却十分精妙.其衍生出各种算法,以致于占据了数据结构的半壁江山.STL中大名顶顶的关联容器--集合(set).映射(map)便是使用二叉树实现.由于篇幅有限,此处仅作一般介绍(如果想要完全了解二叉树以及其衍生出的各种算法,恐怕要写8~10篇). 1)二叉树(Binary Tree) 顾名思义,就是一个节点分出两个节点,称其为左右子节点:每个子节点又可以分出两个子节点,这样递归分叉,其形状很像一颗倒着的树.二叉树限制了每个节点最多有两个子节…

这里演示的二叉树为3层. 递归实现,先构造出一个root节点,先判断左子节点是否为空,为空则构造左子节点,否则进入下一步判断右子节点是否为空,为空则构造右子节点. 利用层数控制迭代次数. 依次递归第二段的内容. 下面是代码,很简单,耐心看看就懂了. package Construct; public class ConstructTree { private int count = 0; class Node { int i; Node left; Node right; public Node…

简单的通过一个寻找嫌疑人的小程序 来演示二叉树的使用 #include <stdio.h> #include <stdlib.h> #include <string.h> /** * 数据结构 - 二叉树 - 节点 */ typedef struct node { char *querstion; struct node *no; struct node *yes; }node; /** * 方法 输入和输入 * 根据打印,询问答案是否正确,y是正确 */ int ye…

题目如下: 题目给出的例子不太好,容易让人误解成不断顺着右节点访问就好了,但是题目意思并不是这样. 换成通俗的意思:按层遍历二叉树,输出每层的最右端结点. 这就明白时一道二叉树层序遍历的问题,用一个队列来处理,但是问题是怎么来辨别每层的最右端结点,我思考了半天,最后想出的办法是利用一个标记位,例如上面的例子: q代表队列,f代表标记结点,right代表记录的最右端结点 q: 1 flag right:{} q: flag 2 3 遇到标记位所以移动标记位,并将队头弹出的数据存起来如下 q: 2…

基于python的list实现二叉树 #!/usr/bin/env python # -*- coding:utf-8 -*- class BinTreeValueError(ValueError): pass class BinTreeList(object): def __init__(self, data, left = None, right = None): self.btree = [data, left, right] #判断二叉树是否为空 def is_empty_bintree…

You are given a binary tree in which each node contains an integer value. Find the number of paths that sum to a given value. The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to…

Given a binary tree, find all leaves and then remove those leaves. Then repeat the previous steps until the tree is empty. Example: Given binary tree 1 / \ 2 3 / \ 4 5 Returns [4, 5, 3], [2], [1]. Explanation: 1. Remove the leaves [4, 5, 3] from the…

One way to serialize a binary tree is to use pre-oder traversal. When we encounter a non-null node, we record the node's value. If it is a null node, we record using a sentinel value such as #. _9_ / \ 3 2 / \ / \ 4 1 # 6 / \ / \ / \ # # # # # # For…

Given a binary tree, return the vertical order traversal of its nodes' values. (ie, from top to bottom, column by column). If two nodes are in the same row and column, the order should be from left to right. Examples: Given binary tree [3,9,20,null,n…

Given a binary tree, find the length of the longest consecutive sequence path. The path refers to any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The longest consecutive path need to be from p…

Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another comput…

Given a binary tree, return all root-to-leaf paths. For example, given the following binary tree: 1 / \ 2 3 \ 5 All root-to-leaf paths are: ["1->2->5", "1->3"] 这道题给我们一个二叉树,让我们返回所有根到叶节点的路径,跟之前那道Path Sum II 二叉树路径之和之二很类似,比那道稍微简单一…

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree. According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w…

Invert a binary tree. 4 / \ 2 7 / \ / \ 1 3 6 9 to 4 / \ 7 2 / \ / \ 9 6 3 1 Trivia: This problem was inspired by this original tweet by Max Howell: Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tre…

Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom. For example:Given the following binary tree, 1 <--- / \ 2 3 <--- \ \ 5 4 <--- You should return [1, 3,…

Given a binary tree where all the right nodes are either leaf nodes with a sibling (a left node that shares the same parent node) or empty, flip it upside down and turn it into a tree where the original right nodes turned into left leaf nodes. Return…

Given a binary tree, return the postorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [3,2,1]. Note: Recursive solution is trivial, could you do it iteratively? 经典题目,求二叉树的后序遍历的非递归方法,跟前序,中序,层序一样都需要用到栈,后续的…

Given a binary tree, find the maximum path sum. The path may start and end at any node in the tree. For example:Given the below binary tree, 1 / \ 2 3 Return 6. 这道求二叉树的最大路径和是一道蛮有难度的题,难就难在起始位置和结束位置可以为任意位置,我当然是又不会了,于是上网看看大神们的解法,看了很多人的都没太看明白,最后发现了网友Yu's…

Given a binary tree, flatten it to a linked list in-place. For example,Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 click to show hints. Hints: If you notice carefully in the flattened tree, each node's right…

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum. For example:Given the below binary tree and sum = 22, 5 / \ 4 8 / / \ 11 13 4 / \ / \ 7 2 5 1 return [ [5,4,11,2], [5,8,4,5] ] 这道二叉树路径之和在之前的基础上又需要找…

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum. For example:Given the below binary tree and sum = 22, 5 / \ 4 8 / / \ 11 13 4 / \ \ 7 2 1 return true…

Given a binary tree, find its minimum depth. The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node. 二叉树的经典问题之最小深度问题就是就最短路径的节点个数,还是用深度优先搜索DFS来完成,万能的递归啊...请看代码: /** * Definition for binary tre…