1119 Pre- and Post-order Traversals (30 分)

Suppose that all the keys in a binary tree are distinct positive integers. A unique binary tree can be determined by a given pair of postorder and inorder traversal sequences, or preorder and inorder traversal sequences. However, if only the postorder and preorder traversal sequences are given, the corresponding tree may no longer be unique.

Now given a pair of postorder and preorder traversal sequences, you are supposed to output the corresponding inorder traversal sequence of the tree. If the tree is not unique, simply output any one of them.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤ 30), the total number of nodes in the binary tree. The second line gives the preorder sequence and the third line gives the postorder sequence. All the numbers in a line are separated by a space.

Output Specification:

For each test case, first printf in a line Yes if the tree is unique, or No if not. Then print in the next line the inorder traversal sequence of the corresponding binary tree. If the solution is not unique, any answer would do. It is guaranteed that at least one solution exists. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input 1:

7
1 2 3 4 6 7 5
2 6 7 4 5 3 1

Sample Output 1:

Yes
2 1 6 4 7 3 5

Sample Input 2:

4
1 2 3 4
2 4 3 1

Sample Output 2:

No
2 1 3 4

题目大意:假设一棵二叉树的所有关键字均是不同的正整数。给出前序遍历和后序遍历,如果中序是唯一的,那么就输出Yes,并且输出中序遍历;如果不唯一,那么就输出No,并且输出其中一种遍历序列即可。

//啊啊,刚看到这个题就很懵,因为以前只知道前序和后序不能确定一棵二叉树,但是也没往深里想。

//不太会,放弃。

主要是因为没有中根无法区分左右子树,对于给的样例来说:

1 2 3 4 和后序2 4 3 1 它的中根可能有:2 1 3 4和2 1 4 3 两种。4无法确定是3的左节点还是右节点。

代码转自:https://www.liuchuo.net/archives/2484

#include <iostream>
#include <vector>
using namespace std;
vector<int> in, pre, post;
bool unique = true;
void getIn(int preLeft, int preRight, int postLeft, int postRight) {
if(preLeft == preRight) {
in.push_back(pre[preLeft]);
return;
}
if (pre[preLeft] == post[postRight]) {//如果找到了根的话。
int i = preLeft + ;
//在前序遍历种找到下一个根节点。
while (i <= preRight && pre[i] != post[postRight-]) i++;
if (i - preLeft > )//如果相当于一个子树为空?它下一个遍历又是根节点?
getIn(preLeft + , i - , postLeft, postLeft + (i - preLeft - ) - );
//最后一个元素是加上当前要去遍历的部分的长度。
else
unique = false;
in.push_back(post[postRight]);//把根push了进去。
getIn(i, preRight, postLeft + (i - preLeft - ), postRight - );
//一次处理之后就把根和左子树
}
}
int main() {
int n;
scanf("%d", &n);
pre.resize(n), post.resize(n);
for (int i = ; i < n; i++)
scanf("%d", &pre[i]);
for (int i = ; i < n; i++)
scanf("%d", &post[i]);
getIn(, n-, , n-);
printf("%s\n%d", unique == true ? "Yes" : "No", in[]);
for (int i = ; i < in.size(); i++)
printf(" %d", in[i]);
printf("\n");
return ;
}

//好难的这道题,不太会,还是需要复习。

PAT 1119 Pre- and Post-order Traversals [二叉树遍历][难]的更多相关文章

  1. PAT 1020 Tree Traversals[二叉树遍历]

    1020 Tree Traversals (25)(25 分) Suppose that all the keys in a binary tree are distinct positive int ...

  2. PAT 甲级 1020 Tree Traversals (二叉树遍历)

    1020. Tree Traversals (25) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Suppo ...

  3. hdu1710(Binary Tree Traversals)(二叉树遍历)

    Binary Tree Traversals Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/O ...

  4. 数据结构作业——order(二叉树遍历)

    order Description 给出一棵二叉树的中序遍历和每个节点的父节点,求这棵二叉树的先序和后 序遍历. Input 输入第一行为一个正整数 n 表示二叉树的节点数目, 节点编号从 1 到 n ...

  5. Construct a tree from Inorder and Level order traversals

    Given inorder and level-order traversals of a Binary Tree, construct the Binary Tree. Following is a ...

  6. leetcode 题解:Binary Tree Level Order Traversal (二叉树的层序遍历)

    题目: Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to ri ...

  7. HDU 1710 Binary Tree Traversals (二叉树遍历)

    Binary Tree Traversals Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/O ...

  8. C++ 二叉树遍历实现

    原文:http://blog.csdn.net/nuaazdh/article/details/7032226 //二叉树遍历 //作者:nuaazdh //时间:2011年12月1日 #includ ...

  9. poj2255 (二叉树遍历)

    poj2255 二叉树遍历 Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu   Descripti ...

随机推荐

  1. iOS swift跑马灯滚动可以点击

    跑马灯,从右至左循环滚动显示信息,并且支持点击事件,使用swift4.0语法完成,更加简介,通用性强,布局部分全部使用snpkit 代码: // // HXQMarqueeView.swift // ...

  2. (译)Getting Started——1.3.3 Working with Foundation(使用Foundation框架)

    在你使用Objective-C语言开发应用时,你会发现在开发中,你会用到很多框架.尤其是Foundation框架,该框架为应用提供了最基础的服务.Foundation框架包括了代表着基本数据类型的va ...

  3. 一图总结C++中关于指针的那些事

    指向对象的指针.指向数据成员的指针,指向成员函数的指针: 数组即指针,数组的指针,指针数组: 指向函数的指针,指向类的成员函数的指针,指针作为函数參数,指针函数: 指针的指针,指向数组的指针:常指针. ...

  4. shell程序练习

    #!/bin/sh通知采用bash解释.如果在echo语句中执行shell命令date,则需要在date命令前面加符号&: 编辑完文件之后不能立即执行该文件,需要给文件设置可执行权限.chmo ...

  5. pom打包参数选择

    pom.xml配置 <profiles> <profile> <id>dev</id> <properties> <token> ...

  6. [Tips]Fix node.js addon build error: "gyp: binding.gyp not found"

    基于node-gyp写Node.js native addon的时候,碰到一个很恶心的问题. 调用“node-gyp configure”能成功,再调用“node-gyp”时总会报错,最后发现时系统时 ...

  7. bootstrap 模式对话框

    <!doctype html> <html> <head> <meta charset="utf-8"> <title> ...

  8. PHP中钩子函数的实现与认识

    PHP中钩子函数的实现与认识 分类:PHP编程  作者:rming  时间:2014-09-21 假如有这么一段程序: function fun(){ fun1(); fun2(); }   首先程序 ...

  9. Vector、ArrayList、List使用深入剖析

    线性表,链表,哈希表是常用的数据结构,在进行Java开发时,JDK已经为我们提供了一系列相应的类来实现基本的数据结构.这些类均在java.util包中.本文试图通过简单的描述,向读者阐述各个类的作用以 ...

  10. 关于mariadb远程连接授权的设置

    1.首先配置允许访问的用户,采用授权的方式给用户权限 1 GRANT ALL PRIVILEGES ON *.* TO 'root'@'%'IDENTIFIED BY '123456' WITH GR ...