[hihoCoder] 后序遍历

The key of this problem is that we need not build the tree from scratch. In fact, we can direct obtain its post-order traversal results in a recursive manner and the problem has given nice hints on this.

The code is as follows.

 #include <iostream>
 #include <string>

 using namespace std;

 string post_order(string pre_order, string in_order) {
     if (pre_order.empty() && in_order.empty())
         return "";
     if (pre_order.length() ==  && in_order.length() == )
         return pre_order;
     string root = pre_order.substr(, );
     int rootInOrder = ;
     while (in_order[rootInOrder] != pre_order[])
         rootInOrder++;
     string pl = pre_order.substr(, rootInOrder);
     string pr = pre_order.substr( + pl.length(), pre_order.length() - pl.length() - );
     string il = in_order.substr(, rootInOrder);
     string ir = in_order.substr(rootInOrder + , in_order.length() - il.length() - );
     return post_order(pl, il) + post_order(pr, ir) + root;
 }

 int main(void) {
     string pre_order, in_order;
     while (cin >> pre_order >> in_order)
         cout << post_order(pre_order, in_order);
     return ;
 }