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;
}