Traversals

HDU (1710)Binary Tree Traversals

A binary tree is a finite set of vertices that is either empty or consists of a root r and two disjoint binary trees called the left and right subtrees. There are three most important ways in which th
·2015-11-03 22:55

03-树2 Tree Traversals Again

这题是第二次做了，两次都不是独立完成，不过我发现我第一次参考的程序，也是参考老师（陈越）的范例做出来的。我对老师给的做了小幅修改，因为我不想有全局变量的的存在，所以我多传了三个参数进去。正序遍历每次都是从1到N吗？看题目我认为应该是，结果我错了，我是对比正确的程序一点点修改才发现的，不容易啊。下面是题目及程序 1 #include <stdio.h> 2 #include &
·2015-11-03 21:00

PAT006 Tree Traversals Again

题目： An inorder binary tree traversal can be implemented in a non-recursive way with a stack. For example, suppose that when a 6-node binary tree (with the keys numbered from 1 to 6) is traversed, the
·2015-11-02 13:54

Tree Traversals

#include<iostream> #include<cstring> #include<queue> using namespace std; //已知二叉树的后序和中序遍历，得到层次遍历顺序。 const int N = 30; struct Node { Node* left; Node* right;
·2015-11-01 16:57

HDU 1710-Binary Tree Traversals（二进制重建）

Binary Tree Traversals Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768
·2015-11-01 11:20

PAT Tree Traversals Again

Tree Traversals Again An inorder binary tree traversal can be implemented in a non-recursive way with
·2015-10-31 19:53

HDU Binary Tree Traversals

#include<stdio.h> #include<cstdlib> int N; int preorder[1010]; int inorder[1010]; int find(int root) {     int i;     for(i=1;i<=N;i++)  &nbs
·2015-10-31 13:08

企业搜索引擎开发之连接器connector（十四）

void startScheduler() { traversalScheduler = (TraversalScheduler) getRequiredBean("TraversalS
·2015-10-31 10:39

hdu 1710 Binary Tree Traversals

根据一颗二叉树的先序遍历结果和中序遍历结果确定后序遍历结果。 先递归建树，后DFS。代码写了详细的注释~~~~ 至今还不会用指针写数据结构。。只会用结构体模拟。。。   #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; const
·2015-10-31 10:01

Tree Traversals (25)

the problem is from pat,which website is http://pat.zju.edu.cn/contests/pat-a-practise/1020 and the source code is as followed. #include<iostream> #include<cstdlib> #include<queue&
·2015-10-31 10:22

Binary Tree Traversals（HDU1710）二叉树的简单应用

  Binary Tree Traversals Time Limit: 1000/1000 MS (Java/Others)    Memory Limit
·2015-10-31 09:32

Tree Traversals (25)

Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the cor
·2015-10-30 13:34

Tree Traversals (25) ——树的遍历

//题目 通过后续遍历 中序遍历 得出一棵树 ，然后按树的层次遍历打印 PS:以前对于这种用指针的题目是比较头痛的，现在做了一些链表操作后，感觉也不难 先通过后续中序建一棵树，然后通过BFS遍历这棵树 提供测试样例 44 1 3 22 3 1 41010 7 6 9 8 5 4 1 3 27 10 6 2 5 9 8 3 1 4 //题目 通过后续遍历 中序遍历 得出一棵树 ，然
·2015-10-27 13:40

HDU 1710 Binary Tree Traversals （二叉树遍历）

Binary Tree Traversals Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768
·2015-10-27 13:00

HDU 1710 二叉树的遍历 Binary Tree Traversals

Binary Tree Traversals Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768
·2015-10-27 12:16

PAT Advance 1020

Tree Traversals (25) 时间限制 400 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard
·2015-10-23 08:07

5-5 Tree Traversals Again (25分)

Aninorderbinarytreetraversalcanbeimplementedinanon-recursivewaywithastack.Forexample,supposethatwhena6-nodebinarytree(withthekeysnumberedfrom1to6)istraversed,thestackoperationsare:push(1);push(2);push
qq_26437925·2015-10-22 11:00

中国大学MOOC-陈越、何钦铭-数据结构-2015秋 03-树3 Tree Traversals Again (25分)

题目传送门~这道题的本意是给定一颗二叉树的先序和中序序列，如何得到这颗二叉树的后序遍历序列，这个问题已经很古老了，然而今天是第一次写，想通了遍历的原理，很容易写出递归版本的求后序序列的函数，至于先序和中序可以从给定的数据获得，即push代表就是先序，pop代表的是中序，不懂请call陈越姥姥。先给出递归版本求后序遍历的递归函数版本。 voidBuild(int*pre,int*in,intn) {
just_sort·2015-09-26 16:00

Tree Traversals

非递归中序遍历Push的顺序为先序遍历Pop的顺序给出中序遍历SampleInput:  6Push1Push2Push3PopPopPush4PopPopPush5Push6PopPopvoidsolve(intpreL,intinL,intpostL,intn)    {if(n==0)return;    if(n==1){post[postL]=pre[preL];return;}    
zhenzhuangde·2015-09-15 21:41

Tree Traversals Again (25)

题目链接：http://www.patest.cn/contests/pat-a-practise/1086题目：Aninorderbinarytreetraversalcanbeimplementedinanon-recursivewaywithastack.Forexample,supposethatwhena6-nodebinarytree(withthekeysnumberedfrom1t
Apie_CZX·2015-09-08 22:00

Tree Traversals (25)多权最短路，和单权是一样的

1020.TreeTraversals(25) 时间限制400ms内存限制65536kB代码长度限制16000B判题程序Standard作者CHEN,YueAtraveler'smapgivesthedistancesbetweencitiesalongthehighways,togetherwiththecostofeachhighway.Nowyouaresupposedtowriteapr
sinat_29278271·2015-08-24 04:00

Tree Traversals (25) - 已知后序和中序求二叉树

1020.TreeTraversals(25)题目地址Supposethatallthekeysinabinarytreearedistinctpositiveintegers.Giventhepostorderandinordertraversalsequences,youaresupposedtooutputthelevelordertraversalsequenceofthecorrespo
qq_26437925·2015-08-13 20:00

Tree Traversals Again (25)

题目如下：Aninorderbinarytreetraversalcanbeimplementedinanon-recursivewaywithastack.Forexample,supposethatwhena6-nodebinarytree(withthekeysnumberedfrom1to6)istraversed,thestackoperationsare:push(1);push(2)
xyt8023y·2015-08-12 14:00

Tree Traversals Again (25)

03-树3.TreeTraversalsAgain(25)时间限制200ms内存限制65536kB代码长度限制8000B判题程序Standard作者CHEN,YueAninorderbinarytreetraversalcanbeimplementedinanon-recursivewaywithastack.Forexample,supposethatwhena6-nodebinarytree(
ice_camel·2015-07-22 09:00

hdu 1710 Binary Tree Traversals 前序遍历和中序推后序

题链;http://acm.hdu.edu.cn/showproblem.php?pid=1710BinaryTreeTraversalsTimeLimit:1000/1000MS(Java/Others)    MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):4205    AcceptedSubmission(s):1904Pro
u013532224·2015-07-06 22:00

Tree Traversals (25) -BFS

题目如下：Supposethatallthekeysinabinarytreearedistinctpositiveintegers.Giventhepostorderandinordertraversalsequences,youaresupposedtooutputthelevelordertraversalsequenceofthecorrespondingbinarytree.InputS
xyt8023y·2015-05-30 13:00

Tree Traversals Again (25)

根据姥姥提示，入栈顺序就是先序遍历，出栈顺序就是中序遍历。然后在用solve函数进行递归分治来求出后序遍历的结果。#include #include #include #include #include #include #include #include #include #include #include #include #include usingnamespacestd; intpre[
xinag578·2015-05-08 21:00

hdu 1710 Binary Tree Traversals

/************************************************Author:xryzEmail:[email protected]:4-2211:41:40FileName:BinaryTreeTraversals.cpp*************************************************/ #include
xinag578·2015-04-22 12:00

Tree Traversals Again (25)及解题材料

03-树2.TreeTraversalsAgain(25)时间限制200ms内存限制65536kB代码长度限制8000B判题程序Standard作者CHEN,YueAninorderbinarytreetraversalcanbeimplementedinanon-recursivewaywithastack.Forexample,supposethatwhena6-nodebinarytree(
tzh476·2015-04-11 20:00

PAT A1086 Tree Traversals Again

//最后一个测试点答案错误，不知道错在哪里了,4分 //又忘记了usingnamespacestd; //用scanf()根本不需要清空！直接每次scanf()就行了 //用string好清空。注意string一定要用cin输入！ #include #include #include #include //#include //#defineLOCAL usingnamespacestd; in
daisyleedq·2015-04-06 22:00

PAT A1020 Tree Traversals 根据先序和中序求层序

//不会增加监视 //空格打不出来，全局变量不能放到局部函数中进行比较、赋值等操作！因为传递的是值，而不是位置。可是书上直接那样写的啊，郁闷 //不是queue*q;是指针队列，队列中存放的就是指针，所以是queueq;第一个是定义一个指向存放node类型的队列的指针 //注意判断是大于等于，重建时递归条件 #include #include #include //#defineLOCAL us
daisyleedq·2015-03-26 23:00

Tree Traversals Again

这个题让我想起“递归改写成循环”这个问题，所有递归都可以用步骤标记的方式改写成循环，尤其是树的非递归式遍历挺有意思#include #include usingnamespacestd; structnode{intlch,rch;}; nodev[60]; intpop[60],id[60]; voidpost(intk){ if(!k)return; staticintfirst=0; pos
u013827143·2015-03-18 11:00

Tree Traversals Again (25)

题目：Aninorderbinarytreetraversalcanbeimplementedinanon-recursivewaywithastack.Forexample,supposethatwhena6-nodebinarytree(withthekeysnumberedfrom1to6)istraversed,thestackoperationsare:push(1);push(2);p
Yangsongtao1991·2015-03-02 19:00

Tree Traversals (25)

Supposethatallthekeysinabinarytreearedistinctpositiveintegers.Giventhepostorderandinordertraversalsequences,youaresupposedtooutputthelevelordertraversalsequenceofthecorrespondingbinarytree.InputSpecif
oFengWuYu1·2015-03-02 17:00

Tree Traversals (25)

题目：Supposethatallthekeysinabinarytreearedistinctpositiveintegers.Giventhepostorderandinordertraversalsequences,youaresupposedtooutputthelevelordertraversalsequenceofthecorrespondingbinarytree.InputSpe
Yangsongtao1991·2015-01-30 22:00

Pat(Advanced Level)Practice--1086(Tree Traversals Again)

Pat1086代码题目描述：Aninorderbinarytreetraversalcanbeimplementedinanon-recursivewaywithastack.Forexample,supposethatwhena6-nodebinarytree(withthekeysnumberedfrom1to6)istraversed,thestackoperationsare:push(1
u012736084·2015-01-09 23:00

hdu1710 Binary Tree Traversals

    题意：给出二叉树的先序和中序遍历，输出它的后序遍历。    思路：对于每个子树，先序的第一个节点肯定是根节点。我们可以在中序中找到这个根节点，该节点左边的长度就是根的左子树的大小，右边同理。然后递归左右子树就可以了。#include #include #include #include #include #include #include #include #include #includ
squee_spoon·2014-09-29 21:00

HDU-1701 Binary Tree Traversals

http://acm.hdu.edu.cn/showproblem.php?pid=1710已知先序和中序遍历，求后序遍历二叉树。思路：先递归建树的过程，后后序遍历。BinaryTreeTraversalsTimeLimit:1000/1000MS(Java/Others)    MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):344
·2014-09-23 20:00

Tree Traversals Again (25)

1086.TreeTraversalsAgain(25)时间限制200ms内存限制32000kB代码长度限制16000B判题程序Standard作者CHEN,YueAninorderbinarytreetraversalcanbeimplementedinanon-recursivewaywithastack.Forexample,supposethatwhena6-nodebinarytree(
solin205·2014-09-09 16:00

Tree Traversals Again

别人的解决方法，感觉蛮优雅的。要加强代码能力啊！代码：#include #include usingnamespacestd; intn,cnt=0; boolfirst=true; structNode { Node(int_val):val(_val),left(NULL),right(NULL){} intval; Node*left; Node*right; }; Node*bui
u014674776·2014-09-09 16:00

PAT 1086 Tree Traversals Again

PAT1086TreeTraversalsAgain题目：Aninorderbinarytreetraversalcanbeimplementedinanon-recursivewaywithastack.Forexample,supposethatwhena6-nodebinarytree(withthekeysnumberedfrom1to6)istraversed,thestackopera
Boostable·2014-09-07 10:00

Tree Traversals Again (25)

pat1086.TreeTraversalsAgain(25)时间限制200ms内存限制32000kB代码长度限制16000B判题程序Standard作者CHEN,YueAninorderbinarytreetraversalcanbeimplementedinanon-recursivewaywithastack.Forexample,supposethatwhena6-nodebinarytr
nan327347465·2014-09-06 21:00

Basic Graph Traversals

Thissectionwillpresentbasicgraphtraversalsbywayofexamplesonthesimplepropertygraphdiagrammedbelow.gremlin>g=TinkerGraphFactory.createTinkerGraph() ==>tinkergraph[vertices:6edges:6] gremlin>v=g.v(1) ==>
macyang·2014-09-03 22:00

HDU 1710 Binary Tree Traversals(二叉树）

题目地址：HDU1710已知二叉树先序和中序求后序。#include #include inta[1001],cnt; typedefstructnode { intdate; node*lchild,*rchild; }*tree; intgetk(intch,intino[],intis,intn) { for(inti=is;idate=pre[ps]; intk=getk(pre[ps],
u013013910·2014-07-31 10:00

Tree Traversals

遵循分治法来写就能过#include #include #include usingnamespacestd; constintN=35; intA[N],B[N],n; structnode{ intval; node*lch,*rch; node(intk):val(k){lch=rch=nullptr;} }; //A[i..j]B[m..n] voidbuild(node*&p,inti,
u013827143·2014-06-25 14:00

Pat(Advanced Level)Practice--1020(Tree Traversals)

Pat1020代码题目描述：Supposethatallthekeysinabinarytreearedistinctpositiveintegers.Giventhepostorderandinordertraversalsequences,youaresupposedtooutputthelevelordertraversalsequenceofthecorrespondingbinarytr
u012736084·2014-04-14 21:00

Tree Traversals (25)

1020.TreeTraversals(25)时间限制400ms内存限制32000kB代码长度限制16000B判题程序Standard作者CHEN,YueSupposethatallthekeysinabinarytreearedistinctpositiveintegers.Giventhepostorderandinordertraversalsequences,youaresupposedt
IAccepted·2014-03-04 17:00

Tree Traversals (25)

链接：http://pat.zju.edu.cn/contests/pat-a-practise/1020   题意：给定二叉树的后序遍历和中序遍历，求层序遍历。   分析：根据后续遍历和中序遍历，可以递归的构建树，建好树之后利用queue层序遍历。   #include<stdio.h> #include<stdlib.h>
249326109·2014-02-27 22:00

Tree Traversals

题目：http://pat.zju.edu.cn/contests/pat-a-practise/1020题解：题意就是给一个二叉树的后序和中序，然后确定这棵二叉树，然后层序输出这棵树。主要思路就是通过后序去确定根root，因为后序中根root肯定是在最后一个的，然后通过找到的根root去把中序的划分成两部分，左边的是根root的左子树，右边的是根root的右子树，再递归左子树和右子树。最终确定二
acm_ted·2014-02-25 22:38

Tree Traversals

题目：http://pat.zju.edu.cn/contests/pat-a-practise/1020题解：题意就是给一个二叉树的后序和中序，然后确定这棵二叉树，然后层序输出这棵树。主要思路就是通过后序去确定根root，因为后序中根root肯定是在最后一个的，然后通过找到的根root去把中序的划分成两部分，左边的是根root的左子树，右边的是根root的右子树，再递归左子树和右子树。最终确定二
ACM_Ted·2014-02-25 22:00
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