从第一个顶点到最后一个Java递归地进行深度优先搜索

Question

i have an assignment in which we have an array of vertices, each vertex has an array list with adjacent vertices. the goal is to use find a path from the first vertex to the last recursively. Once the depth-first search algorithm reaches the destination vertex it adds it to the solution path(doubly linked list), and then recursively adds all the vertices on the direct path back to the source vertex. here is my code so far(instead of solutionPath.add im just printing to console to see what would be added to the linked list with the path)

private DoublyLinkedList<Vertex> dfs(int firstRoom, int lastRoom, boolean[] 
visited){  
 System.out.println(firstRoom);
 if(visited[lastRoom]== true){


    return pathSolution;


}

Iterator<Edge> n = rooms[firstRoom].getEdgesIterator();
while(n.hasNext()){
    int e= n.next().getAdjacentVertex();

    if(!visited[e]){
         visited[e]=true;
        return dfs(e, lastRoom, visited);   

    }


}   


return pathSolution;


 }

Answer 1

You can do it like this:

List dfs(v):
    visited[v] = True
    if v == last_vertex:
        // If v is the last vertex, we can simply return the list containing it
        return [last_vertex]
    for u in neighbors of v:
        if not visited[u]:
             list_from_u = dfs(u)
             // If the list is non-empty, last_vertex is reachable from v
             // and the list contains a valid path to it
             // So we just need to add v to the front of the list and return 
             // the result. Otherwise, there's no path from u to the last_vertex
             // so we do nothing right here and keep looking 
             if not list_from_u.empty():
                  list_from_u.prepend(v)
                  return list_from_u
    // There're was no path to the last_vertex, so we return an empty list
    // to indicate it     
    return []