POJ 2255 Tree Recovery

Tree Recovery

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7819		Accepted: 4947

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

/*
1、前序遍历的第一个字母必是 根
2、在中序遍历的字母串中找出 根字母，那么根字母左右两边的字符串就分别是它的左、右子树
下边是参考别人的代码写的AC代码，有一点不懂就是：
为什么那个构造二叉树的函数是有返回值的，但是调用的时候直接用了 没有将返回的指针赋给任何变量，谁给解释下？
*/
//题意：给出二叉树的先序和中序遍历，让求其后序遍历
#include <iostream>
#include<cstring>
using namespace std;

struct node
{
    char d;
    node* lc;
    node* rc;
}root[50];

node* construct(node* r,char a[],char b[],int n) //构造二叉树
{
    int i;
    if(n == 0) 
        return NULL;
    for (i=0;i<n;i++) 
    {
        if(b[i]==a[0]) 
        {
            r->d=a[0];
            r->lc=construct(&r[1], &a[1],&b[0],i);
            r->rc=construct(&r[i+1], &a[i+1],&b[i+1],n-i-1);
            break;
        }
    }
    return r;
}

void postorder(node* r) //后序遍历
{
    if (r==NULL)
        return;
    postorder(r->lc);
    postorder(r->rc);
    cout<<char(r->d);
//    printf("%c", r->d);
}

int main()
{
    char a[50], b[50];

    while (cin>>a>>b) 
    {
        construct(&root[0],&a[0],&b[0],strlen(a));
        postorder(&root[0]);
        cout<<endl;
    }
    return 0;
}