poj2255Tree Recovery(二叉树，递归)

点击打开原题链接

Tree Recovery

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14027		Accepted: 8760

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 ABCDEFG
BCAD CBAD

Sample Output

ACBFGED
CDAB
题目大意 ：给出二叉树的先序遍历和中序遍历求二叉树的后序遍历。
解题思路：关键是找根，先序遍历的第一个节点就是根，利用这个根可以在中序遍历中将字符分成两个子树，输出根，递归求解。

#include
  
   
#include
   
    
using namespace std;
#define MAX 10000
char a[MAX],b[MAX];
void build(int n,char *a,char *b){
	if(n<=0)return;
	else{
		int p=strchr(b,a[0])-b;
		build(p,a+1,b);
		build(n-1-p,a+p+1,b+p+1);
		printf("%c",a[0]);
	}
}
int main(){
	
	while(scanf("%s %s",a,b)!=EOF){
		int n=strlen(a);
		build(n,a,b);
		printf("\n");
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
	}
}