Topological sort using DFS without recursion

Question

I know the common way to do a topological sort is using DFS with recursion. But how would you do it using stack instead of recursion? I need to obtai

Answer 1

Here we go again. :-) I am submitting an answer, because I do not have enough points to make comments. :-(

Well let me say that I like this algorithm a lot. If the graph is defined in the right way, then there is no error. But take this graph:

vector> graph
{
     { 2, 1 }
    ,{ 2 }
    ,{ }
};

This will display: 2 1 2 0

To protect yourself from graphs defined in that way or where the edges that come in are random you can do this:

#include 
#include 
#include 
#include 
using namespace std;

int main()
{
    stack> dfs;
    stack postorder;
    vector newvector;
    vector::iterator it;
    vector> graph
    {
         { 2, 1 }
        ,{ 2 }
        ,{ }
    };

    vector visited(graph.size());
    vector finallyvisited(graph.size());

    for (int i = 0;i < graph.size();i++)
    {
        if (!visited[i])
        {
            dfs.push(make_pair(false, i));
        }

        while (!dfs.empty())
        {
            pair node = dfs.top();
            dfs.pop();
            if (node.first)
            {
                if (!finallyvisited[node.second])
                {
                    finallyvisited[node.second] = true;
                    postorder.push(node.second);
                    cout << node.second << endl;
                }
                continue;
            }

            visited[node.second] = true;
            dfs.push(make_pair(true, node.second));
            newvector = graph[node.second];
            for (it = newvector.begin();it != newvector.end();it++)
            {
                int son = *it;
                if (!visited[son])
                {
                    dfs.push(make_pair(false, son));
                }
            }
        }
    }
    return 0;
}

Or you could preorder the graph, maybe someone could show that solution. How to preorder randomly given edges that there is no need for second check. :-)

And I did though over the Atif Hussain comment and it is erroneous. That would never work. You always want to push down the stack a node as late as possible so it pops as first as possible.