Tree Recovery

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9710		Accepted: 6105

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 转：

#include 
     
      using namespace std;char pre[100];//先序遍历char in[100];//中序遍历char post[100];//后序遍历int len;//节点个数void solve(int p1,int p2,int m1,int m2){    if(p1>p2)        return ;    int i;    for(i=m1;i<=m2;i++)    {        if(in[i]==pre[p1])            break;    }    post[--len]=pre[p1];//根，放到后序遍历的最后面    if(p1==p2)//叶子节点        return ;    solve(p1+i-m1+1,p2,i+1,m2);//递归处理右子树，得到右子树后序遍历    solve(p1+1,p1+i-m1,m1,i-1);//处理左子树，得到左子树后序遍历}int main(){    while (cin>>pre>>in)    {        memset(post,0,sizeof(post));        len=strlen(pre);        solve(0,len-1,0,len-1);        cout<
      
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