Найти кратчайший путь рекурсивно, принимая во внимание родительский узел в графе, хранящемся как словарь

Question

У меня есть график, как показано ниже:

Я сохраняю его в структуре данных словаря.Предположим, я хочу найти путь от D до M:

Затем я хочу что-то вроде поиска кратчайшего пути от D до M:

DHKM

, а затем взять его родительские узлы, гдеон находит путь от AM без D в нем,

AEILM

и BM без D в нем,

BEILM

На данный момент я смогнайти только кратчайший путь от DM.

class Graph(object):

    def __init__(self, graph_dict=None):
        """ initializes a graph object
            If no dictionary or None is given,
            an empty dictionary will be used
        """
        if graph_dict == None:
            graph_dict = {}
        self.__graph_dict = graph_dict

    def vertices(self):
        """ returns the vertices of a graph """
        return list(self.__graph_dict.keys())

    def edges(self):
        """ returns the edges of a graph """
        return self.__generate_edges()

    def add_vertex(self, vertex):
        """ If the vertex "vertex" is not in
            self.__graph_dict, a key "vertex" with an empty
            list as a value is added to the dictionary.
            Otherwise nothing has to be done.
        """
        if vertex not in self.__graph_dict:
            self.__graph_dict[vertex] = []

    def add_edge(self, edge):
        """ assumes that edge is of type set, tuple or list;
            between two vertices can be multiple edges!
        """
        edge = set(edge)
        (vertex1, vertex2) = tuple(edge)
        if vertex1 in self.__graph_dict:
            self.__graph_dict[vertex1].append(vertex2)
        else:
            self.__graph_dict[vertex1] = [vertex2]

    def __generate_edges(self):
        """ A static method generating the edges of the
            graph "graph". Edges are represented as sets
            with one (a loop back to the vertex) or two
            vertices
        """
        edges = []
        for vertex in self.__graph_dict:
            for neighbour in self.__graph_dict[vertex]:
                if {neighbour, vertex} not in edges:
                    edges.append({vertex, neighbour})
        return edges

    def __str__(self):
        res = "vertices: "
    for k in self.__graph_dict:
        res += str(k) + " "
        res += "\nedges: "
        for edge in self.__generate_edges():
            res += str(edge) + " "
        return res

    def find_path(self, start_vertex, end_vertex, path=[]):
        """ find all paths from start_vertex to
            end_vertex in graph """
        graph = self.__graph_dict
        path = path + [start_vertex]
        if start_vertex == end_vertex:
            return [path]
        if start_vertex not in graph:
            return []
        paths = []
        for vertex in graph[start_vertex]:
            if vertex not in path:
                extended_paths = self.find_path(vertex,
                                                 end_vertex,
                                                 path)
                for p in extended_paths:
                    paths.append(p)
        return paths


graph_dict = { 'a' : ['d', 'e'],
      'b' : ['d', 'e'],
      'c' : ['h'],
      'd' : ['a', 'b', 'h'],
      'e' : ['a', 'b', 'i'],
      'f' : ['i'],
      'g' : ['k'],
      'h' : ['c', 'd', 'k'],
      'i' : ['e', 'f', 'l'],
      'j' : ['l'],
      'k' : ['g', 'h', 'm'],
      'l' : ['i', 'j', 'm'],
      'm' : ['k', 'l']
    }

graph = Graph(graph_dict)


print('The path from vertex "d" to vertex "m":')
path = graph.find_path('d', 'm')
min = len(path[0])
for path_list in path:
    if len(path_list)<min:
        min = path_list
print(min)

Спасибо!