POJ-2255-Tree Recovery-求后序

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

题意：给了二叉树的前序、中序，求后序

思路：

前序遍历：根-->左孩子-->右孩子
中序遍历：左孩子-->根-->右孩子
后序遍历：左孩子-->右孩子-->根
所谓的前中后指的是根的位置，而左右孩子顺序是不变的。

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。
以此类推递归下去。

递归找到，代码很简单，但是要明白怎么递归：

 参考博客：http://www.cnblogs.com/-sunshine/archive/2012/07/24/2606341.html

学到一个新函数：

strchr：用来找s2中首次出现s1[0]的位置，返回位置指针，若不存在则返回NULL

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

void print(int n,char *s1,char *s2,char *s)
{
    if(n<=0)
        return;
    int x=strchr(s2,s1[0])-s2;
    //strchr用来找s2中首次出现s1[0]的位置，返回位置指针，若不存在则返回NULL
    print(x,s1+1,s2,s);//得到左子树
    print(n-1-x,s1+x+1,s2+x+1,s+x);//得到右子树
    s[n-1]=s1[0];//最后一个是根
}
int main()
{
    char s1[30],s2[30],s3[30];
    memset(s1,'\0',sizeof(s1));
    memset(s2,'\0',sizeof(s2));
    memset(s3,'\0',sizeof(s3));
    while(~scanf("%s %s",s1,s2))
    {
        print(strlen(s1),s1,s2,s3);
        s3[strlen(s1)]='\0';//没有这个输出CDABGED不是CDAB
        printf("%s\n",s3);
    }
    return 0;
}