python3.x - 关于Python图遍历的操作

Question

就是创建了一个图 想要进行深度遍历 和 广度遍历 但是第二个遍历的时候只会出现一个data 感觉是因为自己之前的那个遍历把 self.visited[node] = True 的缘故 但是又不知道怎么进行修改，求各位指教

以下是代码：

class Graph(object):
    def __init__(self, *args, **kwargs):
        self.node_neighbors = {}
        self.visited = {}

    def add_nodes(self,nodelist):
        for node in nodelist:
            self.add_node(node)

    def add_node(self,node):
        if node not in self.nodes():
            self.node_neighbors[node] = []

    def add_edge(self,edge):
        u, v = edge
        if(v not in self.node_neighbors[u]) and (u not in self.node_neighbors[v]):
            self.node_neighbors[u].append(u)
            if(u!=v):
                self.node_neighbors[v].append(u)

    def nodes(self):
        return self.node_neighbors.keys()

    def depth_first_search(self, root=None):
        order = []
        def dfs(node):
            self.visited[node] = True
            order.append(node)
            for  n in self.node_neighbors[node]:
                if not n in self.visited:
                    dfs(n)
        if root:
            dfs(root)
        for node in self.nodes():
            if not node in self.visited:
                dfs(node)
        print(order)
        return order

    def breadtg_frist_search(self, root = None):
        queue = []
        order = []
        def bfs():
            while len(queue) >  0:
                node = queue.pop()
                self.visited[node] = True
                for n in self.node_neighbors[node]:
                    if (not n in self.visited) and (not n in queue):
                        queue.append(n)
                        order.append(n)
        if root:
            queue.append(root)
            order.append(root)
            bfs()
        for node in self.nodes():
            if not node in self.visited:
                queue.append(node)
                order.append(node)
                bfs()
        print(order)
        return order


if __name__ == '__main__':
    g = Graph()
g.add_nodes([i+1 for i in range(10)])
g.add_edge((1, 2))
g.add_edge((1, 3))
g.add_edge((2, 4))
g.add_edge((2, 5))
g.add_edge((4, 8))
g.add_edge((5, 8))
g.add_edge((5, 9))
g.add_edge((3, 6))
g.add_edge((3, 7))
g.add_edge((7, 10))
g.add_edge((9, 10))
print('nodes:',  g.nodes())
order = g.depth_first_search(1)
order = g.breadtg_frist_search(1)

然后遍历出来的结果是

nodes: dict_keys([1, 2, 3, 4, 5, 6, 7, 8, 9, 10])
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
[1]

Answer 1

楼主，是self.visited的问题,第一次深度搜索调用self.visted时,已经把所有节点变为true,第二次广度搜索使用第一次深度搜索结果,　改为如下即可:

class Graph(object):
    def __init__(self, *args, **kwargs):
        self.node_neighbors = {}
#        self.visited = {}    #　删除此行
    ...

    def depth_first_search(self, root=None):
        self.visited = {}    # 添加此行
        ...

    def breadtg_frist_search(self, root = None):
        self.visited = {}    # 添加此行
        ...