How to implement depth first search for graph with a non-recursive approach

Question

I have spent lots of time on this issue. However, I can only find solutions with non-recursive methods for a tree: Non recursive for tree, or a recursive method for the grap

Answer 1

Using Stack and implementing as done by the call stack in the recursion process-

The Idea is to push a vertex in the stack, and then push its vertex adjacent to it which is stored in a adjacency list at the index of the vertex and then continue this process until we cannot move further in the graph, now if we cannot move ahead in the graph then we will remove the vertex which is currently on the top of the stack as it is unable to take us on any vertex which is unvisited.

Now, using stack we take care of the point that the vertex is only removed from the stack when all the vertices that can be explored from the current vertex have been visited, which was being done by the recursion process automatically.

for ex -

See the example graph here.

( 0 ( 1 ( 2 ( 4 4 ) 2 ) ( 3 3 ) 1 ) 0 ) ( 6 ( 5 5 ) ( 7 7 ) 6 )

The above parenthesis show the order in which the vertex is added on the stack and removed from the stack, so a parenthesis for a vertex is closed only when all the vertices that can be visited from it have been done.

(Here I have used the Adjacency List representation and implemented as a vector of list (vector > AdjList) by using C++ STL)

void DFSUsingStack() {
    /// we keep a check of the vertices visited, the vector is set to false for all vertices initially.
    vector visited(AdjList.size(), false);

    stack st;

    for(int i=0 ; i :: iterator it = AdjList[curr].begin() ; it != AdjList[curr].end() ; it++){
                if(visited[*it] == false){
                    st.push(*it);
                    cout << (*it) << '\n';
                    visited[*it] = true;
                    break;
                }
            }
            /// We can move ahead from current only if a new vertex has been added on the top of the stack.
            if(st.top() != curr){
                continue;
            }
            st.pop();
        }
    }
}