tree（nyoj）

Tree

时间限制：1000 ms | 内存限制：65535 KB

难度：3

描述

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!

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

// 输入前缀和中缀写出后缀式， 两种方法//利用递归通过依次遍历前序来组织中序
#include <stdio.h>
#include <string.h>
#include <stdlib.h>#define MAXLEN 27void rebuid_binary_tree(char *pre_str, char *in_str, int len)
{char ch;if(len == 0)                                            //该递归的结束条件为该元素的左右两边都没有元素可以继续return ;ch = *pre_str;int i;for(i = 0; i < len; i++){if(in_str[i] == *pre_str)break;}                                               rebuid_binary_tree(pre_str + 1, in_str, i);                               //左子树rebuid_binary_tree(pre_str + i + 1, in_str + i +1, len - i - 1);       //右子树printf("%c", ch);
}int main()
{char pre_str[MAXLEN], in_str[MAXLEN];while(scanf("%s %s", pre_str, in_str) != EOF){int len = strlen(pre_str);rebuid_binary_tree(pre_str, in_str, len);printf("\n");}return 0;
} // 将字符串分割为三部分
#include<stdio.h>
#include<string.h>char pre_str[27], in_str[27];                       //两个字符串设为全局变量void rcreat(int p_s, int p_e, int i_s, int i_e);    //重建一棵树int main()
{int len;while(scanf("%s%s", pre_str, in_str) != EOF){len = strlen(pre_str) - 1;rcreat(0, len, 0, len); printf("\n");}return 0;
}void rcreat(int p_s, int p_e, int i_s, int i_e)
{if(p_e < p_s)                  // 没有一边的子树的时return;                                                  //两个if都是递归结束的条件if(p_s == p_e)                 //子树只有一个元素时结束{printf("%c", in_str[i_s]);                   //中序和后序都是左子树开始遍历return;}int i;for(i = i_s; ;i++){if(pre_str[p_s] == in_str[i])break;}int len = i - i_s;rcreat(p_s + 1, p_s +len, i_s, i - 1);         //     把相应的字符串传进来rcreat(p_s + len + 1, p_e, i + 1, i_e);printf("%c", in_str[i]);
}

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