HLG2040 two fork tree traversal known before, after seeking

Traversal of a binary tree

Time limit:1000 MS Memory limit:32768 K Total submit:60(users) Total accepted:34(users) Rating: Special Judge: No

Description

A binary tree is given a middle order and a pre-sequence traversal, which outputs its post-order traversal.

Input

The subject has multiple sets of data, input processing to the end of the file. The first row of each set of data includes an integer n, which indicates that the binary tree has n nodes in total. The next line contains n integers for each row, indicating the middle sequence traversal of this tree. The next line contains n integers for each row, indicating the first sequence traversal of the tree. 3<= N <= 100

Output

Each set of outputs includes a row that indicates the sequential traversal of this tree.

Sample Input

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

Sample Output

4 5 2 6 7 3 1

Source

Winter Holiday Contest 5

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace Std;

int n,num=0;
int KMP (int root,int str2[])
{
int s=0;
while (1)
{
if (str2[s]==root)
return s;
s++;
}
return 0;
}
void calculate (int m,int str1[],int str2[],int str3[])
{
if (m<=0)
return;
int LEN=KMP (STR1[0],STR2);
Calculate (LEN,STR1+1,STR2,STR3);
Calculate (M-LEN-1,STR1+LEN+1,STR2+LEN+1,STR3);
STR3[NUM++]=STR1[0];
}
int main ()
{
int n;
int str1[105],str2[105],str3[105];
while (scanf ("%d", &n)!=eof)
{
num=0;
memset (str1,0,sizeof (str1));
memset (str2,0,sizeof (str2));
memset (str3,0,sizeof (STR3));
for (int i=0;i<n;i++)
scanf ("%d", &str2[i]);
for (int i=0;i<n;i++)
scanf ("%d", &str1[i]);
Calculate (N,STR1,STR2,STR3);
for (int i=0;i<n;i++)
printf ("%d", str3[i]);
printf ("\ n");
}
return 0;
}

HLG2040 two fork tree traversal known before, after seeking