即使这是一个有点老的问题，我还是想为遇到问题的任何人提供一个切实可行的答案。

假设您以如下的元组列表的形式获取连接的输入数据：

[('A', 'B'), ('B', 'C'), ('B', 'D'), ('C', 'D'), ('E', 'F'), ('F', 'C')]

我发现对Python中的图形最有用和最有效的数据结构是 集合的决定
。这将是我们Graph班级的基础结构。您还必须知道这些连接是弧形（定向，以一种方式连接）还是边缘（无定向，以两种方式连接）。我们将通过directed向该Graph.__init__方法添加参数来处理该问题。我们还将添加一些其他有用的方法。

import pprint
from collections import defaultdict


class Graph(object):
    """ Graph data structure, undirected by default. """

    def __init__(self, connections, directed=False):
        self._graph = defaultdict(set)
        self._directed = directed
        self.add_connections(connections)

    def add_connections(self, connections):
        """ Add connections (list of tuple pairs) to graph """

        for node1, node2 in connections:
            self.add(node1, node2)

    def add(self, node1, node2):
        """ Add connection between node1 and node2 """

        self._graph[node1].add(node2)
        if not self._directed:
            self._graph[node2].add(node1)

    def remove(self, node):
        """ Remove all references to node """

        for n, cxns in self._graph.items():  # python3: items(); python2: iteritems()
            try:
                cxns.remove(node)
            except KeyError:
                pass
        try:
            del self._graph[node]
        except KeyError:
            pass

    def is_connected(self, node1, node2):
        """ Is node1 directly connected to node2 """

        return node1 in self._graph and node2 in self._graph[node1]

    def find_path(self, node1, node2, path=[]):
        """ Find any path between node1 and node2 (may not be shortest) """

        path = path + [node1]
        if node1 == node2:
            return path
        if node1 not in self._graph:
            return None
        for node in self._graph[node1]:
            if node not in path:
                new_path = self.find_path(node, node2, path)
                if new_path:
                    return new_path
        return None

    def __str__(self):
        return '{}({})'.format(self.__class__.__name__, dict(self._graph))

我将其作为创建读者find_shortest_path和其他方法的“读者练习” 。

让我们来看看实际情况…

>>> connections = [('A', 'B'), ('B', 'C'), ('B', 'D'),
                   ('C', 'D'), ('E', 'F'), ('F', 'C')]
>>> g = Graph(connections, directed=True)
>>> pretty_print = pprint.PrettyPrinter()
>>> pretty_print.pprint(g._graph)
{'A': {'B'},
 'B': {'D', 'C'},
 'C': {'D'},
 'E': {'F'},
 'F': {'C'}}

>>> g = Graph(connections)  # undirected
>>> pretty_print = pprint.PrettyPrinter()
>>> pretty_print.pprint(g._graph)
{'A': {'B'},
 'B': {'D', 'A', 'C'},
 'C': {'D', 'F', 'B'},
 'D': {'C', 'B'},
 'E': {'F'},
 'F': {'E', 'C'}}

>>> g.add('E', 'D')
>>> pretty_print.pprint(g._graph)
{'A': {'B'},
 'B': {'D', 'A', 'C'},
 'C': {'D', 'F', 'B'},
 'D': {'C', 'E', 'B'},
 'E': {'D', 'F'},
 'F': {'E', 'C'}}

>>> g.remove('A')
>>> pretty_print.pprint(g._graph)
{'B': {'D', 'C'},
 'C': {'D', 'F', 'B'},
 'D': {'C', 'E', 'B'},
 'E': {'D', 'F'},
 'F': {'E', 'C'}}

>>> g.add('G', 'B')
>>> pretty_print.pprint(g._graph)
{'B': {'D', 'G', 'C'},
 'C': {'D', 'F', 'B'},
 'D': {'C', 'E', 'B'},
 'E': {'D', 'F'},
 'F': {'E', 'C'},
 'G': {'B'}}

>>> g.find_path('G', 'E')
['G', 'B', 'D', 'C', 'F', 'E']