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 stack operations are: push(1); push(2); push(3); pop(); pop(); push(4); pop(); pop(); push(5); push(6); pop(); pop(). Then a unique binary tree (shown in Figure 1) can be generated from this sequence of operations. Your task is to give the postorder traversal sequence of this tree.


Figure 1

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (≤) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 2 lines follow, each describes a stack operation in the format: "Push X" where X is the index of the node being pushed onto the stack; or "Pop" meaning to pop one node from the stack.

Output Specification:

For each test case, print the postorder traversal sequence of the corresponding tree in one line. A solution is guaranteed to exist. All the numbers must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input:

6
Push 1
Push 2
Push 3
Pop
Pop
Push 4
Pop
Pop
Push 5
Push 6
Pop
Pop

Sample Output:

3 4 2 6 5 1
#include<cstdio>
#include<cstring>
#include<stack>
using namespace std;
const int maxn = ;
struct Node{
int data;
Node* lchild;
Node* rchild;
}; int pre[maxn],in[maxn];
int n,num = ; Node* createTree(int preL,int preR,int inL,int inR){
if(preL > preR) return NULL;
Node* root = new Node;
root->data = pre[preL];
int k;
for(k = inL; k <= inR; k++){
if(in[k] == pre[preL]) break;
}
int numLeft = k - inL;
root->lchild = createTree(preL+,preL+numLeft,inL,k-);
root->rchild = createTree(preL+numLeft+,preR,k+,inR);
return root;
} void postOrder(Node* root){
if(root == NULL) return;
postOrder(root->lchild);
postOrder(root->rchild);
printf("%d",root->data);
num++;
if(num < n) printf(" ");
} int main(){
scanf("%d",&n);
char str[];
int x,preIndex = ,inIndex = ;
stack<int> s;
for(int i = ; i < *n; i++){
getchar();
scanf("%s",str);
if(strcmp(str,"Push") == ){
scanf("%d",&x);
s.push(x);
pre[preIndex++] = x;
}else{
in[inIndex++] = s.top();
s.pop();
}
}
Node* root = createTree(,n-,,n-);
postOrder(root);
return ;
}

03-树3 Tree Traversals Again (25 分)的更多相关文章

  1. PTA 03-树3 Tree Traversals Again (25分)

    题目地址 https://pta.patest.cn/pta/test/16/exam/4/question/667 5-5 Tree Traversals Again   (25分) An inor ...

  2. PAT 甲级 1086 Tree Traversals Again (25分)(先序中序链表建树,求后序)***重点复习

    1086 Tree Traversals Again (25分)   An inorder binary tree traversal can be implemented in a non-recu ...

  3. 数据结构课后练习题(练习三)7-5 Tree Traversals Again (25 分)

    7-5 Tree Traversals Again (25 分)   An inorder binary tree traversal can be implemented in a non-recu ...

  4. 【PAT甲级】1086 Tree Traversals Again (25 分)(树知二求一)

    题意:输入一个正整数N(<=30),接着输入2*N行表示栈的出入(入栈顺序表示了二叉搜索树的先序序列,出栈顺序表示了二叉搜索树的中序序列),输出后序序列. AAAAAccepted code: ...

  5. A1020 Tree Traversals (25 分)

    Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and i ...

  6. PAT A1020 Tree Traversals (25 分)——建树,层序遍历

    Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and i ...

  7. 1020 Tree Traversals (25 分)(二叉树的遍历)

    给出一个棵二叉树的后序遍历和中序遍历,求二叉树的层序遍历 #include<bits/stdc++.h> using namespace std; ; int in[N]; int pos ...

  8. 03-树3 Tree Traversals Again (25 分)

    An inorder binary tree traversal can be implemented in a non-recursive way with a stack. For example ...

  9. 03-树2. Tree Traversals Again (25)

    03-树2. Tree Traversals Again (25) 时间限制 200 ms 内存限制 65536 kB 代码长度限制 8000 B 判题程序 Standard 作者 CHEN, Yue ...

  10. 03-树3. Tree Traversals Again (25)将先序遍历和中序遍历转为后序遍历

    03-树3. Tree Traversals Again (25) 题目来源:http://www.patest.cn/contests/mooc-ds/03-%E6%A0%913 An inorde ...

随机推荐

  1. mybatis Generator 生成配置文件

    <?xml version="1.0" encoding="UTF-8" ?><!DOCTYPE generatorConfiguration ...

  2. Jenkins执行selenium报错unknown error: cannot find Chrome binary

    问题描述:在Pycharm中执行selenium测试用例,可以正常运行, 集成在Jenkins中,构建时,发现构建成功,但是查看Console Output,报错:unknown error: can ...

  3. 搭建vue脚手架,包含Axios、qs、Element-UI、mock等插件的安装配置

    1.安装node.vue这些最基础最简单的安装的就一一省略过. 1.1 axios 安装 1.2安装 Element-Ui 插件 1.3 安装 qs 1.4  安装 Mock 2.新建一个vue工程, ...

  4. windows server2012怎样关机怎样重启-详细教程

    | 浏览:1991 | 更新:2014-12-15 17:33 1 2 3 4 5 6 分步阅读 百度经验:jingyan.baidu.com windows server2012和以往有些不同,关机 ...

  5. Linux 基础教程 29-tcpdump命令-1

    什么是tcpdump     在Linux中输入命令man tcpdump给出的定义如下所示: tcpdump - 转储网络上的数据流 是不是感觉很懵?我们用通俗.形象.学术的表达方式来全方位描述tc ...

  6. MYSQL - JSON串中查找key对应的值

    1.建表 -- 建表 drop table if exists ta_product2; CREATE TABLE ta_product2( id int primary key auto_incre ...

  7. Firefox mobile (android) and orientationchange

    Firefox for Android does not support the orientationchange event but you can achieve the same result ...

  8. 随手记录: MVC自定义提交form

    function mySubmit() { var frm = $('#frm'); var result = frm.valid(); if (ret) { frm.submit(); } else ...

  9. c# 利用t4模板,自动生成Model类

    我们在用ORM(比如dapper)的时候,很多时候都需要自己写Model层(当然也有很多orm框架自带了这种功能,比如ef),特别是表里字段比较多的时候,一个Model要写半天,而且Model如果用于 ...

  10. Pi 在Windows下面使用远程桌面登录

    1.删除系统自带的xrdp 输入命令sudo apt-get purge xrdp pi@raspberrypi:~ $ sudo apt-get purge xrdp 正在读取软件包列表... 完成 ...