B - Tree Recovery POJ - 2255

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 ABCDEFG
BCAD CBAD

Sample Output

ACBFGED
CDAB

题目大意；就是知道搜索二叉树 前序遍历，中序遍历 求后序遍历；

解题思路：例如已知前序遍历是DBACEGF，中序遍历是ABCDEFG，那么由前序遍历先根，可知道D是树的根，再看在中序遍历中D左边是ABC，所以可知道ABC一定在D的左子树上，而EFG在D的右子树上。
那么前序遍历为BAC,中序遍历为ABC，所以B为根，在中序遍历中A在B的左边，C在B的右边，所以A为B的左孩子，C为B的有孩子。

 

代码

1. #include<stdio.h>
2. #include<string.h>
3. void dfs(int n,char *s1,char *s2)
4. {
5.     if(n<=0) return;
6.     int t=strchr(s2,s1[0])-s2; //找到前序遍历中根节点在中序遍历的位置
7.     
8.     dfs(t,s1+1,s2);            //找所有的左孩子
9.     
10.     dfs(n-t-1,s1+1+t,s2+1+t);  //找右孩子
11.  
12.     printf("%c",s1[0]);   
13. }
14. int main()
15. {
16.     char s1[50],s2[50];
17.     while(~scanf("%s %s",s1,s2))
18.     {
19.         int n=strlen(s1);
20.         dfs(n,s1,s2);
21.         printf("\n");
22.     }
23.     return 0;
24. }