ZOJ Problem Set

Tree Recovery

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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：

已知先序遍历和中序遍历字符串，求后序遍历

解题思路：

已知先序和中序，可以唯一的确定后续

已知后序和中序，可以唯一的确定先序

但是已知先序和后续，不能唯一确定中序

回到题目，先根据先序和中序确定根节点位置，然后用递归来求子树

AC：

#include <cstdio>#include <iostream>#include <string>using namespace std;string strPost;void PostOrder(string pre, string in){int iPos;if(pre == "" || in == "")return;iPos = in.find(pre[0]);PostOrder(pre.substr(1, iPos), in.substr(0, iPos));//substr的用法是返回一个新的string（即子串）,第一位表示起始地址，第二位表示后面的多少位PostOrder(pre.substr(iPos + 1), in.substr(iPos+1));strPost += pre[0];}int main(){string pre,in;while(cin>>pre>>in){strPost = "";PostOrder(pre, in);cout<<strPost<<endl;}return 0;}

完整：

#include <cstdio>#include <iostream>#include <string>using namespace std;string strPost;string strPre;void PostOrder(string pre, string in){int iPos;if(pre == "" || in == "")return;iPos = in.find(pre[0]);PostOrder(pre.substr(1, iPos), in.substr(0, iPos));//substr的用法是返回一个新的string（即子串）,第一位表示起始地址，第二位表示后面的多少位PostOrder(pre.substr(iPos + 1), in.substr(iPos+1));strPost += pre[0];}void PreOrder(string in, string post){long iPos;if(in == "" || post == "")return;iPos = in.find(post[post.size()-1]);strPre += post[post.size()-1];PreOrder(in.substr(0,iPos), post.substr(0, iPos));PreOrder(in.substr(iPos + 1, in.size() - iPos -1), post.substr(iPos, post.size() - iPos - 1) );}int main(){string pre,in,post;while(cin>>pre>>in>>post){strPost = "";PostOrder(pre, in);PreOrder(in, post);cout<<strPost<<endl;cout<<strPre<<endl;}return 0;}