poj2255解題報告

來源:互聯網
上載者:User
Tree Recovery
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 4585   Accepted: 3061

Description

Little Valentine liked playing with binary trees very much. Her favorite game was constructing randomly looking binary trees with capital letters in the nodes.
This is an example of one of her creations:

D
/ /
/ /
B E
/ / /
/ / /
A C G
/
/
F

To record her trees for future generations, she wrote down two strings for each tree: a preorder traversal (root, left subtree, right subtree) and an inorder traversal (left subtree, root, right subtree). For the tree drawn above the preorder traversal is DBACEGF and the inorder traversal is ABCDEFG.
She thought that such a pair of strings would give enough information to reconstruct the tree later (but she never tried it).

Now, years later, looking again at the strings, she realized that reconstructing the trees was indeed possible, but only because she never had used the same letter twice in the same tree.
However, doing the reconstruction by hand, soon turned out to be tedious.
So now she asks you to write a program that does the job for her!

Input

The input will contain one or more test cases.
Each test case consists of one line containing two strings preord and inord, representing the preorder traversal and inorder traversal of a binary tree. Both strings consist of unique capital letters. (Thus they are not longer than 26 characters.)
Input is terminated by end of file.

Output

For each test case, recover Valentine's binary tree and print one line containing the tree's postorder traversal (left subtree, right subtree, root).

Sample Input

DBACEGF ABCDEFGBCAD CBAD

Sample Output

ACBFGEDCDAB
題意:給出二叉樹的兩種遍曆序列,輸出第三種遍曆序列....
思路:昨天資料結構課上老師講的方法,,晚上回來寫的程式....
#include<iostream>using namespace std;struct T{char d;int l,r;}T[30];char str1[30],str2[30],lock[30];int head;void find(int th,int tree){int i,j;for(j=0;j<strlen(str1);j++)for(i=th-1;i>-1&&lock[i]==0;i--)if(str1[j]==str2[i]){T[tree].l=i;lock[i]=1;find(i,i);goto end;}end:for(j=0;j<strlen(str1);j++)for(i=th+1;i<strlen(str1)&&lock[i]==0;i++)if(str1[j]==str2[i]){T[tree].r=i;lock[i]=1;find(i,i);goto end1;}end1:return ;}void s(int tree){if(T[tree].l!=-1)s(T[tree].l);if(T[tree].r!=-1)s(T[tree].r);cout<<T[tree].d;}int main(){int i;while(scanf("%s%s",str1,str2)!=EOF){memset(lock,0,sizeof(lock));for(i=0;i<strlen(str2);i++){T[i].d=str2[i];T[i].l=T[i].r=-1;if(str2[i]==str1[0])head=i;}lock[head]=1;find(head,head);s(head);cout<<endl;}return 0;}

聯繫我們

該頁面正文內容均來源於網絡整理,並不代表阿里雲官方的觀點,該頁面所提到的產品和服務也與阿里云無關,如果該頁面內容對您造成了困擾,歡迎寫郵件給我們,收到郵件我們將在5個工作日內處理。

如果您發現本社區中有涉嫌抄襲的內容,歡迎發送郵件至: info-contact@alibabacloud.com 進行舉報並提供相關證據,工作人員會在 5 個工作天內聯絡您,一經查實,本站將立刻刪除涉嫌侵權內容。

A Free Trial That Lets You Build Big!

Start building with 50+ products and up to 12 months usage for Elastic Compute Service

  • Sales Support

    1 on 1 presale consultation

  • After-Sales Support

    24/7 Technical Support 6 Free Tickets per Quarter Faster Response

  • Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.