[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() == 1 && in_order.length() == 1)
        return pre_order;
    string root = pre_order.substr(0, 1);
    int rootInOrder = 0;
    while (in_order[rootInOrder] != pre_order[0])
        rootInOrder++;
    string pl = pre_order.substr(1, rootInOrder);
    string pr = pre_order.substr(1 + pl.length(), pre_order.length() - pl.length() - 1);
    string il = in_order.substr(0, rootInOrder);
    string ir = in_order.substr(rootInOrder + 1, in_order.length() - il.length() - 1);
    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 0;
}