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, *rchild;
}; int in[maxn] = {}, pre[maxn] = {};
int num = ; Node* createTree(int preL, int preR, int inL, int inR);
void postOrder(Node *root,int n); int main()
{
int n;
scanf("%d",&n); int x;
int preIndex = , inIndex = ;
char str[];
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
{
x = s.top();
s.pop();
in[inIndex++] = x;
}
} Node *root = createTree(,n-,,n-);
postOrder(root,n);
return ;
} 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,int n)
{
if (root == NULL)
{
return;
}
postOrder(root->lchild,n);
postOrder(root->rchild,n);
printf("%d",root->data); num++;
if (num < n)
{
printf(" ");
}
}

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 ...

  9. 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. DotnetSpider爬虫简单示例 net core

    文章地址 https://blog.csdn.net/sD7O95O/article/details/78097556 安装爬虫框架  NUGET 安装DotnetSpider 创建HTTP协议数据包 ...

  2. java基础 抽象

    /** * 抽象方法:就是加上abstract关键字,并去掉大括号,分号结束 * 抽象类:抽象方法坐在的类,必须是抽象类.在class前家abstract即可 * * 如何使用抽象类和抽象方法: * ...

  3. 理解 BLS 签名算法

    理解 BLS 签名算法 来源 https://medium.com/cryptoadvance/bls-signatures-better-than-schnorr-5a7fe30ea716 原文标题 ...

  4. witchcase

    #include "stdafx.h" #include using namespace std; int _tmain(int argc, _TCHAR* argv[]) { i ...

  5. npm err! Unexpected end of JSON input while parsing near解决办法

    npm install时出现npm err! Unexpected end of JSON input while parsing near错误 输入  npm cache clean --fore ...

  6. js学习之数据结构和算法

    js中的数据结构 1.列表 待办事项列表.购物清单.最佳十名榜单等等. 适用: 1)数据结构较为简单, 2)不需要在一个长序列中查找元素,或者对其进行排序 2.栈 一摞盘子 ----- 添加删除只能从 ...

  7. Redis学习总结整理

    基本操作 设值 set KEY VALUE 获值 get KEY 设置过期时间(单位:s) set KEY VALUE ex SECONDS 设置过期时间(单位:s) exprire KEY SECO ...

  8. 关于IE环境下按回车键会自动触发button 按钮的点击事件的解决方案

    今天项目中遇到IE不兼容的问题,于是就根据问题进行修改,修改过程中发现,在输入框内用扫码枪扫描东西后会自动执行页面下面的button按钮,但是其它浏览不会出现这样的问题. 解决方案: 1.用a标签 2 ...

  9. 详解Linux操作系统的进程

    系统 计算机运行起来以后,就是由内核和运行在内核之上的众多进程来实现的(kernel+process) 内存分为 :    线性内存: 物理内存: 计算机的所有运行都只在内存和CPU中运行! 内核空间 ...

  10. C语言几个术语: 数据对象,左值,右值

    1. 数据对象 赋值表达式语句的目的是把值存储到内存位置上. 用于存储值的数据存储区域统称为数据对象. 2. 左值 左值是C语言的术语, 用于标识特定数据对象的名称或表达式. 对象指的是实际的数据存储 ...