dfs_old_practice.py

def dfs(graph, visited, node):
    #if node not in visited:
    print(node)
    if node not in visited:
        visited.append(node)
# you made the mistake of putting the for loop in the same position as the if statement
        for i in graph[node]:
            #print(node)
            #print(i, node)
            dfs(graph, visited, i)
           
    return visited
    # return is executed on the CURRENT FUNCTION CALL
    # It terminates only the CURRENT FUNCTION CALL and goes back up to the prev 
    # function call. 
    # Because the for loop calls keys that is why we had the key error. So need to
    # specify if a key does not have any children.
    # What if you encounter something that was already visited? Then it returns the list
    # Is that what a DFS is? You should ask this.


graph = {'A': ['B','C'], 'B': ['D', 'F'], 'D': ['E','F'], 'E': ['D'],'F': [], 'C': []}
# Does it require everything to be connected? Yes, Probablyy

graph1 = {
    'A' : ['B','S'],
    'B' : ['A'],
    'C' : ['D','E','F','S'],
    'D' : ['C'],
    'E' : ['C','H'],
    'F' : ['C','G'],
    'G' : ['F','S'],
    'H' : ['E','G'],
    'S' : ['A','C','G']
}

def test(): # Problem: I am not entirely sure if this is correct. -> ask someone
    if dfs(graph, [], 'A') == ['A', 'B', 'D', 'E', 'F', 'C']:
        return True
    else:
        return False

#print(dfs(graph, [], 'A'))
print(dfs(graph, [], 'A'))