POJ 2255 Tree Recovery 二叉树基础

Tree Recovery Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10353   Accepted: 6513 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 Source Ulm Local 1997

首先。。。学到了一个不需要建树的方法，利用了遍历的方法进行递归，很简洁：

建树的代码我稍后上传吧

code1：不建树的方法

#include <iostream>
using namespace std;
string s1, s2;

void findpost(int root, int left, int right);
int main()
{
	while (cin>>s1>>s2){
		findpost(0,0,s2.size()-1);
		cout<<endl;
	}
	return 0;
}

void findpost(int root,int left,int right)
{
	if (left>right) return;
	for (int i=left; i<=right && s1[root]!=s2[i];i++);
	findpost(root + 1, left, i - 1);
	findpost(root + i - left + 1, i + 1, right);
	cout << s2[i];
}

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