CLRS Topological sort algorithm reverse of DFS?

Question

I have one confusion looking at the step for topological sort i see that the reverse order of DFS is a prospective solution for Topological sort.

I also tried a small code

void graph::dfs(void)
{
    for(std::vector<vertex>::iterator iter =vertexes.begin();iter < vertexes.end();iter++ ){
        iter->visited = WHITE;  
    }   

    for(std::vector<vertex>::iterator iter =vertexes.begin();iter < vertexes.end();iter++ ){
        if(iter->visited == WHITE){
            dfs_visits(*iter);
        }
    }

    std::cout << "-----------------dfs------------------"<<std::endl;
    for(std::list<int>::reverse_iterator riter = q.rbegin() ; riter != q.rend();riter++)
        std::cout << *riter << std::endl;

    std::cout << "-----------------topological_sort------------------"<<std::endl;
    for(std::list<int>::iterator iter = q.begin() ; iter != q.end();iter++)
        std::cout << *iter << std::endl;


    q.clear();
}


void graph::dfs_visits(vertex& source){
    source.visited = GREY;
    for(std::list<edge>::iterator iter = source.list.begin();iter != source.list.end();iter++){
        if(vertexes[iter->destination_vertex].visited == WHITE){
            dfs_visits(vertexes[iter->destination_vertex]);
        }
    }
    source.visited = BLACK;
    q.push_front(source.id);
}

The graph data structure is here

#include "iostream"
#include "vector"
#include "list"

enum color{
    WHITE,
    GREY,
    BLACK
};

struct edge{
    int destination_vertex;
    edge(int ver){
        destination_vertex = ver;
    }
};

struct vertex{
    int id;
    color visited;
    std::list<edge>  list;
    vertex(int _id){
        id = _id;
    }
};


class graph
{
private:
    std::vector<vertex> vertexes;
    int next;
    std::list<int> q;
public:
    graph(void){
        next = 0;
    }
    ~graph(void){}
    void add_node(std::vector<int> edges );
    void add_node(std::vector<int> incoming_edges , std::vector<int> outgoing_edges);
    void print();
    void dfs();
    void dfs_visits(vertex& source);
    void bfs();
    static void process();
};

Here is one e.g. graph i tried

0->1,2,
1->3,
2->
3->
4->
5->4,
-----------------dfs------------------
3
1
2
0
4
5
-----------------topological_sort-----
5
4
0
2
1
3

Changing the question statement

My question is really simple.. Is topological sort is always DFS in reverse order? If not is there a counter example?

If you see my output for the particular graph the DFS output and its reverse is a correct solution for topological sort of the graph too....also reading the CLR topological sort alorithm it also looks like topological sort is the reverse of DFS?

Answer 1

Yes it's always true. The reverse of a DFS on a DAG gives the topological sort.

Source: http://www.cse.ust.hk/faculty/golin/COMP271Sp03/Notes/MyL08.pdf slide 13