二叉树重建

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:

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 file 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 ABCDEFG
BCAD CBAD
Sample Output CDAB

#include<iostream>#include<cstring>#include<cstdio>using namespace std;void build(int n,char* s1,char* s2,char* s)//递归{    if(n<=0)        return ;  //strchr找到字符在字符串中第1次出现的位置    int  p=strchr(s2,s1[0])-s2;//找到根节点在中序遍历中的位置    build(p,s1+1,s2,s);//递归构造左子树的后序遍历    build(n-p-1,s1+p+1,s2+p+1,s+p);//递归构造右子树的后序遍历    s[n-1]=s1[0];//把根节点添加到最后}int main(){    char s1[30],s2[30],ans[30];    while(~scanf("%s%s",s1,s2))    {        int n=strlen(s1);        build(n,s1,s2,ans);        ans[n]='\0';//字符串结束标志        printf("%s\n",ans);    }    return 0;}