nyoj221 Tree

                                                D

                                              / 
                                             /   
                                            B     E

                                           / \     
                                          /   \     
                                         A     C     G

                                                    /

                                                   /

                                                  F

输入

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.

输出

For each test case, recover Valentine‘s binary tree and print one line containing the tree‘s postorder traversal (left subtree, right subtree, root).

样例输入

DBACEGF ABCDEFG
BCAD CBAD

样例输出

ACBFGED
CDAB

思路：先说下本题的大意：有多组测试数据，已知先序和中序，求后序

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
typedef struct node{

	char data;
	struct node*lchild,*rchild;
}NODE,*PNODE;
PNODE rebuildTree(char *post,char*in,int len)//post表示先序遍历的结果，in表示中序遍历的结果，len表示当前子树的长度
{
    if(len<=0)
		return NULL;
	PNODE node=(PNODE)malloc(sizeof(NODE));
	node->data=*post;
	char* k=strchr(in,node->data);//找到根节点所在的中序的位置
	int leftlen=strlen(in)-strlen(k);//左子树长度
	int rightlen=len-leftlen-1;//右子树长度
	node->lchild=rebuildTree(post+1,in,leftlen);//递归建立左子树
	node->rchild=rebuildTree(post+leftlen+1,in+leftlen+1,rightlen);//递归建立右子树
	return node;
}
void print(PNODE root)//后序遍历输出结果
{
	if(root==NULL)
		return;
	print(root->lchild);
	print(root->rchild);
	printf("%c",root->data);
}
int main()
{
	char post[27],in[27];
	while(~scanf("%s%s",post,in))
	{
		PNODE root=NULL;
	   root=rebuildTree(post,in,strlen(in));
       print(root);
	   printf("\n");
	}
	return 0;
}