二叉树已知前序中序求后序

#include <string>
#include <iostream>
using namespace std;

string FindPostOrder( string pre_order, string in_order );

int main()
{
	string pre_order, in_order, post_order;
	cin >> pre_order >> in_order;
	cout << FindPostOrder( pre_order, in_order );
	system("pause");
	return 0;
}

string FindPostOrder( string pre_order, string in_order )
{
	if( pre_order.length() != in_order.length() )   //前序中序元素个数不相等出错
	{
		cout << "Failed" << endl;
		exit(1);
	}

	if( pre_order.length() == 1 )                //递归终止
	{
		if( pre_order != in_order )               //一个元素时前序中序不相等
		{
			cout << "Failed" <<endl;
			exit(1);
		}
		return pre_order;
	}

	if( pre_order.length() == 0)                  //前序为空时，后序也为空
		return string();

	char node = pre_order[0];					//取第一个元素
	int index = in_order.find( node );			//找到其在中序中的位置

	if( index == string::npos )					//没有找到，则出错
	{
		cout << "Failed" <<endl;
		exit(1);
	}
	int len = pre_order.length();

	//左子树的后序
	string left_part = FindPostOrder( pre_order.substr(1,index), in_order.substr(0,index) ); 
	//右子树的后序
	string right_part = FindPostOrder( pre_order.substr(1+index, len-index-1), in_order.substr(1+index,len-index-1) );

	return left_part+right_part+node; //串联起来

}