POJ 2255/递归：前序中序求后序

Sample Input

DBACEGF ABCDEFG

已知二叉树的前序遍历与后序遍历求后序遍历

算法：前序遍历时，第一位为根：D；找到中序中的D，则前面的ABC在左子树，右边的EFG在右子树，则后序为为左子树+右子树+根:solve(BAC,ABC)+solve(EGF,EFG)+D

#include <iostream>

#include <string>

using namespace std;

string solve(string pre,string mid){

	if(pre.length()==1)return pre;

	else if(pre.length()==0)return "";

	int m = mid.find(pre[0]);

	

	return solve(pre.substr(1,m),mid.substr(0,m))+solve(pre.substr(m+1),mid.substr(m+1))+pre[0];

}

int main(int argc, char* argv[])

{

	

	string pre,mid;

	while(cin>>pre>>mid){

		

		cout<<solve(pre,mid)<<endl;

	}

	return 0;

}