已知树的前序和中序求后序 hdu 题目1710 Binary Tree Traversals
来源:互联网 发布:java前沿技术 2017 编辑:程序博客网 时间:2024/06/05 05:49
Binary Tree Traversals
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2448 Accepted Submission(s): 1066
Problem Description
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 the vertices of a binary tree can be systematically traversed or ordered. They are preorder, inorder and postorder. Let T be a binary tree with root r and subtrees T1,T2.
In a preorder traversal of the vertices of T, we visit the root r followed by visiting the vertices of T1 in preorder, then the vertices of T2 in preorder.
In an inorder traversal of the vertices of T, we visit the vertices of T1 in inorder, then the root r, followed by the vertices of T2 in inorder.
In a postorder traversal of the vertices of T, we visit the vertices of T1 in postorder, then the vertices of T2 in postorder and finally we visit r.
Now you are given the preorder sequence and inorder sequence of a certain binary tree. Try to find out its postorder sequence.
In a preorder traversal of the vertices of T, we visit the root r followed by visiting the vertices of T1 in preorder, then the vertices of T2 in preorder.
In an inorder traversal of the vertices of T, we visit the vertices of T1 in inorder, then the root r, followed by the vertices of T2 in inorder.
In a postorder traversal of the vertices of T, we visit the vertices of T1 in postorder, then the vertices of T2 in postorder and finally we visit r.
Now you are given the preorder sequence and inorder sequence of a certain binary tree. Try to find out its postorder sequence.
Input
The input contains several test cases. The first line of each test case contains a single integer n (1<=n<=1000), the number of vertices of the binary tree. Followed by two lines, respectively indicating the preorder sequence and inorder sequence. You can assume they are always correspond to a exclusive binary tree.
Output
For each test case print a single line specifying the corresponding postorder sequence.
Sample Input
91 2 4 7 3 5 8 9 64 7 2 1 8 5 9 3 6
Sample Output
7 4 2 8 9 5 6 3 1
利用递归,从前序一个个元素开始,m=1; 在中序中找到in[i] == pre[m], 然后将中序分为两部分,in[1.....i-1] 和in[i+1.....n],然后分别遍历这两部分,直到这两部分元素为0(s>t) 或1个(s==t);
#include <iostream>#include <stdio.h>#include <stdlib.h>using namespace std;int m = 1;int p = 1;int pre[1003],in[1003],post[1003];void Traverse(int s,int t){ if(s>t) return;//空,没有元素 if(s==t){//一个元素 m++; post[p++] = in[s]; // printf("%d ",in[s]); return ; } int k,i; //printf("---------pre[%d]=%d\n",m,pre[m]); k = pre[m++]; i= s; while(in[i++]!=k);// printf("i=%d\n",i); Traverse(s,i-2); Traverse(i,t); post[p++] = in[i-1];// printf("%d ",in[i-1]); }/*91 2 4 7 3 5 8 9 64 7 2 1 8 5 9 3 6*/int main(){ int n,i; while(scanf("%d",&n)!=EOF) { for(i=1;i<=n;i++) scanf("%d",&pre[i]); for(i=1;i<=n;i++) scanf("%d",&in[i]); // for(i=1;i<=n;i++) printf("%d ",pre[i]); // for(i=1;i<=n;i++) printf("%d ",in[i]); Traverse(1,n); // printf("\n"); for(i=1;i<n;i++) printf("%d ",post[i]); printf("%d\n",post[n]);//这里格式问题!!! m=1; p=1; //每组测试数据都要将m,p初始化!!! } return 0;}
- 已知树的前序和中序求后序 hdu 题目1710 Binary Tree Traversals
- HDU 1710 Binary Tree Traversals(已知先序中序求后序)
- hdoj 1710 Binary Tree Traversals 【二叉树由前序和中序求后序】
- hdu 1710 Binary Tree Traversals 前序遍历和中序推后序
- HDU 1710 Binary Tree Traversals 二叉树
- HDU 1710 Binary Tree Traversals(二叉树)
- hdu 1710 Binary Tree Traversals 二叉树的遍历
- HDU 1710(Binary Tree Traversals)二叉树的遍历
- HDU 1710 - Binary Tree Traversals(树的遍历)
- HDU 1710-Binary Tree Traversals
- HDU 1710 Binary Tree Traversals
- HDU--1710 -- Binary Tree Traversals
- hdu--1710Binary Tree Traversals
- hdu 1710 Binary Tree Traversals
- Hdu--1710--Binary Tree Traversals
- HDU-1710-Binary Tree Traversals
- hdu-1710-Binary Tree Traversals
- HDU 1710 Binary Tree Traversals
- S3C2416 Linux2.6.21 驱动移植--添加UART3 及波特率设置bug消除
- 胆识也是一种能力[转 一个女程序员的创业人生]
- 浅谈Java 同步机制
- UVa 10651 Pebble Solitaire 记忆化搜索+位运算
- uva 10025 The ? 1 ? 2 ? ... ? n = k problem(算式规律)
- 已知树的前序和中序求后序 hdu 题目1710 Binary Tree Traversals
- join & union & group by
- wince 关于Bin文件的解析
- Maya 2013 下载以及破解教程
- Objective-C之Foundation框架NSNumber、NSNull、NSValue、NSDate用法介绍
- Hibernate之session
- Eboot中nand flash控制器参数TACLS、TWRPH0和TWRPH1的确定(基于K9F2G08U0B)
- UVA 10591 Happy Number
- Windows多线程编程总结