Python 实现广度优先搜索

Question

我在网上找到了一个例子，但是，仅返回 BFS 元素的序列不足以进行计算。假设根是 BFS 树的第一级，然后它的子级是第二级，等等。我怎么知道它们在哪个级别，以及从下面的代码中每个节点的父级是谁(我将创建一个对象存储其父级和树级别)？

# sample graph implemented as a dictionary
graph = {'A': ['B', 'C', 'E'],
         'B': ['A','D', 'E'],
         'C': ['A', 'F', 'G'],
         'D': ['B'],
         'E': ['A', 'B','D'],
         'F': ['C'],
         'G': ['C']}

# visits all the nodes of a graph (connected component) using BFS
def bfs_connected_component(graph, start):
   # keep track of all visited nodes
   explored = []
   # keep track of nodes to be checked
   queue = [start]

   # keep looping until there are nodes still to be checked
   while queue:
       # pop shallowest node (first node) from queue
       node = queue.pop(0)
       if node not in explored:
           # add node to list of checked nodes
           explored.append(node)
           neighbours = graph[node]

           # add neighbours of node to queue
           for neighbour in neighbours:
               queue.append(neighbour)
   return explored

bfs_connected_component(graph,'A') # returns ['A', 'B', 'C', 'E', 'D', 'F', 'G']

Answer 1

您可以通过首先将级别 0 分配给起始节点来跟踪每个节点的级别。然后对于节点 X 的每个邻居分配级别 level_of_X + 1 .

此外，您的代码多次将同一个节点插入队列。我使用了一个单独的列表 visited为了避免这种情况。

# sample graph implemented as a dictionary
graph = {'A': ['B', 'C', 'E'],
         'B': ['A','D', 'E'],
         'C': ['A', 'F', 'G'],
         'D': ['B'],
         'E': ['A', 'B','D'],
         'F': ['C'],
         'G': ['C']}


# visits all the nodes of a graph (connected component) using BFS
def bfs_connected_component(graph, start):
    # keep track of all visited nodes
    explored = []
    # keep track of nodes to be checked
    queue = [start]

    levels = {}         # this dict keeps track of levels
    levels[start]= 0    # depth of start node is 0

    visited= [start]     # to avoid inserting the same node twice into the queue

    # keep looping until there are nodes still to be checked
    while queue:
       # pop shallowest node (first node) from queue
        node = queue.pop(0)
        explored.append(node)
        neighbours = graph[node]

        # add neighbours of node to queue
        for neighbour in neighbours:
            if neighbour not in visited:
                queue.append(neighbour)
                visited.append(neighbour)

                levels[neighbour]= levels[node]+1
                # print(neighbour, ">>", levels[neighbour])

    print(levels)

    return explored

ans = bfs_connected_component(graph,'A') # returns ['A', 'B', 'C', 'E', 'D', 'F', 'G']
print(ans)