传送门1: https://uva.onlinejudge.org/external/5/536.pdf 传送门2: http://acm.gdufe.edu.cn/Problem/read/id/1077 题意一样   输入不一样 HINT: 1. Preorder : (root, left subtree, right subtree); Inorder : (left subtree, root, right subtree); Postorder: (left subtree, rig…

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:                                               …

题意:给出一颗二叉树的前序遍历和中序遍历,输出其后序遍历 用杭电1710的代码改一点,就可以了. #include<iostream> #include<cstdio> #include<cstring> #include <cmath> #include<algorithm> using namespace std; typedef long long LL; ; char a[maxn],b[maxn],ans[maxn],s1[maxn],…

题意: 给出先序和中序,求后序. 思路: ①建树然后递归输出. //建树的 #include<iostream> #include<cstdio> #include<queue> #include<cstring> using namespace std; int n; ],mid[]; ]; struct node{ char c; node *left,*right; node() { c='a'; left=right=NULL; } }; ,coun…

题意:已知先序中序,输出后序. #pragma comment(linker, "/STACK:102400000, 102400000") #include<cstdio> #include<cstring> #include<cstdlib> #include<cctype> #include<cmath> #include<iostream> #include<sstream> #include&…

题目 题目     分析 莫名A了     代码 #include <bits/stdc++.h> using namespace std; string s1,s2; void build(int l1,int r1,int l2,int r2) { int root=l1,p=l2; if(l1>r1) return; while(s2[p]!=s1[root] && p<=r2) p++; int cnt=p-l2; build(l1+1,l1+cnt,l2,…

Tree Recovery Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11939   Accepted: 7493 Description Little Valentine liked playing with binary trees very much. Her favorite game was constructing randomly looking binary trees with capital le…

 Tree Recovery  Little Valentine liked playing with binary trees very much. Her favoritegame was constructingrandomly looking binary trees with capital letters in the nodes. This is an example of one of her creations: D / \ / \ B E / \ \ / \ \ A C G…

Tree Recovery Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 14640   Accepted: 9091 Description Little Valentine liked playing with binary trees very much. Her favorite game was constructing randomly looking binary trees with capital le…

Tree Recovery Description 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 / \ \…

UVA.548 Tree(二叉树 DFS) 题意分析 给出一棵树的中序遍历和后序遍历,从所有叶子节点中找到一个使得其到根节点的权值最小.若有多个,输出叶子节点本身权值小的那个节点. 先递归建树,然后DFS求解. 代码总览 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <string> #include <sstre…

Tree Recovery POJ - 2255 根据树的前序遍历和中序遍历还原后序遍历. (偷懒用了stl的find) #include<iostream> #include<string> using namespace std; string s1,s2; int len; void work(int l1,int r1,int l2,int r2) { if(l1==r1) { cout<<s1[l1]; return; } if(l1>r1) retur…

Tree Recovery 时间限制:C/C++ 1秒,其他语言2秒 空间限制:C/C++ 131072K,其他语言262144K 64bit IO Format: %lld 链接:https://www.nowcoder.com/acm/contest/77/H 题目描述:进行区间求和,区间加减(线段树加懒惰标记) 输入 10 5 1 2 3 4 5 6 7 8 9 10 Q 4 4 Q 1 10 Q 2 4 C 3 6 3 Q 2 4 输出 4 55 9 15 #include<cstdio…

[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 递归题 [代码] #include <bits/stdc++.h> using namespace std; const int N = 300; string s1, s2; int n, idx; int g[N][2]; int dfs(int l, int r) { int temp = l; for (int i = l; i <= r; i++) { if (s2[i] == s1[idx]) { temp…

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: To record her trees for future generations, sh…

先声明,我还在学习中,这个题大部分代码借鉴的大佬的,其实这算是比较经典二叉树题了,关键在于递归建树. 代码附上: #include <iostream> #include <cstring> using namespace std; ]; ]; void TL( int p1, int p2, int q1, int q2, int root ) { if ( p1 > p2 ) return; for ( root = q1 ; TM[root] != TF[p1] ; +…

题目来源:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=3&page=show_problem&problem=48 Tree Summing  Background LISP was one of the earliest high-level programming languages and, with FORTRAN, is one of the olde…

You are to determine the value of the leaf node in a given binary tree that is the terminal node of a path of least value from the root of the binary tree to any leaf. The value of a path is the sum of values of nodes along that path. Input The input…

题意: 计算从根到叶节点的累加值,看看是否等于指定值.是输出yes,否则no.注意叶节点判断条件是没有左右子节点. 思路: 建树过程中计算根到叶节点的sum. 注意: cin读取失败后要调用clear恢复,否则后面无法正常读取. 注意空树都要输出no 最初代码如下:   #include<cstdio> #include<cstring> #include<iostream> #include<string> #include<algorithm>…

前序和中序输入二叉树,后序输出二叉树:核心思想只有一个,前序的每个根都把中序分成了两部分,例如 DBACEGF ABCDEFG D把中序遍历的结果分成了ABC和EFG两部分,实际上,这就是D这个根的左右子树,而接下来的B,把ABC分成了A和C两部分,那么,A,C实际上是B的左右子树而E把EFG分成了空集和FG两部分,也就是说,E只有右子树,F把FG分成了空集和G两部分,即F只有右子树G.个人认为理解了思路以后,最重要的细节就是遍历 前序遍历 的结果时,你的那个p的增加(这里的P指的是要把前序遍历…

题目来源: http://poj.org/problem?id=2255 题目描述: Description 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 cr…

版权声明:本文为博主原创文章,未经博主同意不得转载.vasttian https://blog.csdn.net/u012860063/article/details/37699219 转载请注明出处:viewmode=contents" rel="nofollow">http://blog.csdn.net/u012860063?viewmode=contents 题目链接:http://poj.org/problem?id=2255 Description Litt…

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 / / F To recor…

内存池技术就是创建一个内存池,内存池中保存着可以使用的内存,可以使用数组的形式实现,然后创建一个空闲列表,开始时将内存池中所有内存放入空闲列表中,表示空闲列表中所有内存都可以使用,当不需要某一内存时,将其放入空闲列表中,使内存可以循环使用.空闲列表可以用不定长数组vector定义,内存池和空闲列表的类型由使用内存的数据决定,如果使用内存的数据是用户自定义的结构体类型,则使用此类型.由于大部分情况下是使用指针进行内存调用,所以空闲列表中存储的是相应的指针. 具体实例如下: 题目: Trees ar…

You are to determine the value of the leaf node in a given binary tree that is the terminal node of a path of least value from the root of the binary tree to any leaf. The value of a path is the sum of values of nodes along that path. Input The input…

Description You are to determine the value of the leaf node in a given binary tree that is the terminal node of apath of least value from the root of the binary tree to any leaf. The value of a path is the sum of valuesof nodes along that path.InputT…

0.这是一道利用中序遍历和后序遍历确定二叉树的题目,学会建树 关键点理解这段代码 int build(int L1,int R1,int L2,int R2) { //printf("built:\n"); ;//空树 int root=post_order[R2]; int p=L1; while(in_order[p] != root) p++; int cnt = p-L1;//左子树的结点个数 lch[root]=build(L1,p-,L2,L2+cnt-); rch[roo…

http://poj.org/problem?id=2255 #include<cstdio> #include <cstring> using namespace std; const int maxn = 27; char pre[maxn],in[maxn]; char past[maxn]; void tre(int ps,int pe,int is,int ie,int& ind) { int lnum = strchr(in,pre[ps]) - in - is…

题意:给出一颗二叉树的前序遍历和中序遍历的序列,让你输出后序遍历的序列. 思路:见代码,采用递归. #include <iostream> #include <stdio.h> #include <string.h> /* 由一颗二叉树的前序遍历和中序遍历,输出该二叉树的后序遍历. */ using namespace std; ; int len; char dlr[maxn],ldr[maxn]; //dlr:前序遍历的序列,ldr:中序遍历的序列 /* l1,r1…

题意:给出一颗点带权的二叉树的中序和后序遍历,找一个叶子使得它到根的路径上的权和最小. 学习的紫书:先将这一棵二叉树建立出来,然后搜索一次找出这样的叶子结点 虽然紫书的思路很清晰= =可是理解起来好困难啊啊啊啊 后来终于问懂一丢丢了--- 比如说样例: 中序遍历:3 2 1 4 5 7 6 后序遍历:3 1 2 5 6 7 4 首先做第一层: 在后序遍历中的最后一个数为根节点,然后在到中序遍历中找到这个根节点,在这个根节点的左边是左子树,右边是右子树,这样就确定出了左子树和右子树的区间 然后做第…