如何在广度优先搜索中跟踪路径？

如何跟踪广度优先搜索的路径，如以下示例所示：

如果搜索key 11，则返回连接1到11 的最短列表。

[1, 4, 7, 11]

您应该先查看http://en.wikipedia.org/wiki/Breadth-first_search。

下面是一个快速实现，其中我使用列表列表表示路径队列。

# graph is in adjacent list representation
graph = {
        '1': ['2', '3', '4'],
        '2': ['5', '6'],
        '5': ['9', '10'],
        '4': ['7', '8'],
        '7': ['11', '12']
        }

def bfs(graph, start, end):
    # maintain a queue of paths
    queue = []
    # push the first path into the queue
    queue.append([start])
    while queue:
        # get the first path from the queue
        path = queue.pop(0)
        # get the last node from the path
        node = path[-1]
        # path found
        if node == end:
            return path
        # enumerate all adjacent nodes, construct a new path and push it into the queue
        for adjacent in graph.get(node, []):
            new_path = list(path)
            new_path.append(adjacent)
            queue.append(new_path)

print bfs(graph, '1', '11')

另一种方法是维护从每个节点到其父节点的映射，并在检查相邻节点时记录其父节点。搜索完成后，只需根据父映射进行回溯即可。

graph = {
        '1': ['2', '3', '4'],
        '2': ['5', '6'],
        '5': ['9', '10'],
        '4': ['7', '8'],
        '7': ['11', '12']
        }

def backtrace(parent, start, end):
    path = [end]
    while path[-1] != start:
        path.append(parent[path[-1]])
    path.reverse()
    return path


def bfs(graph, start, end):
    parent = {}
    queue = []
    queue.append(start)
    while queue:
        node = queue.pop(0)
        if node == end:
            return backtrace(parent, start, end)
        for adjacent in graph.get(node, []):
            if node not in queue :
                parent[adjacent] = node # <<<<< record its parent 
                queue.append(adjacent)

print bfs(graph, '1', '11')

上面的代码基于没有循环的假设。

我非常喜欢qiao的第一个答案！这里唯一缺少的是将顶点标记为已访问。

为什么我们需要这样做？
让我们想象一下，从节点11连接了另一个节点号13。现在我们的目标是找到节点13。
经过一点点运行，队列将如下所示：

[[1, 2, 6], [1, 3, 10], [1, 4, 7], [1, 4, 8], [1, 2, 5, 9], [1, 2, 5, 10]]

请注意，最后有两个路径的节点编号为10。
这意味着从节点号10开始的路径将被检查两次。在这种情况下，它看起来并不那么糟糕，因为10号节点没有任何子节点。但是，这可能真的很糟糕（即使在这里我们也会无故检查两次该节点。）
13号节点不在其中这些路径，因此程序在到达最后一个节点号为10的第二条路径之前不会返回。我们将对其进行重新检查。

我们所缺少的只是一个标记访问的节点而不要再次检查它们的集合。
这是修改后的qiao的代码：

graph = {
    1: [2, 3, 4],
    2: [5, 6],
    3: [10],
    4: [7, 8],
    5: [9, 10],
    7: [11, 12],
    11: [13]
}


def bfs(graph_to_search, start, end):
    queue = [[start]]
    visited = set()

    while queue:
        # Gets the first path in the queue
        path = queue.pop(0)

        # Gets the last node in the path
        vertex = path[-1]

        # Checks if we got to the end
        if vertex == end:
            return path
        # We check if the current node is already in the visited nodes set in order not to recheck it
        elif vertex not in visited:
            # enumerate all adjacent nodes, construct a new path and push it into the queue
            for current_neighbour in graph_to_search.get(vertex, []):
                new_path = list(path)
                new_path.append(current_neighbour)
                queue.append(new_path)

            # Mark the vertex as visited
            visited.add(vertex)


print bfs(graph, 1, 13)

该程序的输出将是：

[1, 4, 7, 11, 13]

没有轻松的检查。

非常简单的代码。每次发现节点时，您都​​会追加路径。

graph = {
         'A': set(['B', 'C']),
         'B': set(['A', 'D', 'E']),
         'C': set(['A', 'F']),
         'D': set(['B']),
         'E': set(['B', 'F']),
         'F': set(['C', 'E'])
         }
def retunShortestPath(graph, start, end):

    queue = [(start,[start])]
    visited = set()

    while queue:
        vertex, path = queue.pop(0)
        visited.add(vertex)
        for node in graph[vertex]:
            if node == end:
                return path + [end]
            else:
                if node not in visited:
                    visited.add(node)
                    queue.append((node, path + [node]))

我以为我会尝试将此代码编写起来很有趣：

graph = {
        '1': ['2', '3', '4'],
        '2': ['5', '6'],
        '5': ['9', '10'],
        '4': ['7', '8'],
        '7': ['11', '12']
        }

def bfs(graph, forefront, end):
    # assumes no cycles

    next_forefront = [(node, path + ',' + node) for i, path in forefront if i in graph for node in graph[i]]

    for node,path in next_forefront:
        if node==end:
            return path
    else:
        return bfs(graph,next_forefront,end)

print bfs(graph,[('1','1')],'11')

# >>>
# 1, 4, 7, 11

如果需要循环，可以添加以下内容：

for i, j in for_front: # allow cycles, add this code
    if i in graph:
        del graph[i]

我既喜欢@乔的第一个答案，又喜欢@Or的加法。为了减少处理量，我想补充一下Or的答案。

在@Or的答案中，访问节点的跟踪很棒。我们还可以允许程序比当前状态早退出。在for循环的某个点上，current_neighbour必须是end，一旦发生，就会找到最短路径，程序可以返回。

我将修改方法如下，请密切注意for循环

graph = {
1: [2, 3, 4],
2: [5, 6],
3: [10],
4: [7, 8],
5: [9, 10],
7: [11, 12],
11: [13]
}


    def bfs(graph_to_search, start, end):
        queue = [[start]]
        visited = set()

    while queue:
        # Gets the first path in the queue
        path = queue.pop(0)

        # Gets the last node in the path
        vertex = path[-1]

        # Checks if we got to the end
        if vertex == end:
            return path
        # We check if the current node is already in the visited nodes set in order not to recheck it
        elif vertex not in visited:
            # enumerate all adjacent nodes, construct a new path and push it into the queue
            for current_neighbour in graph_to_search.get(vertex, []):
                new_path = list(path)
                new_path.append(current_neighbour)
                queue.append(new_path)

                #No need to visit other neighbour. Return at once
                if current_neighbour == end
                    return new_path;

            # Mark the vertex as visited
            visited.add(vertex)


print bfs(graph, 1, 13)

输出和其他所有内容都将相同。但是，该代码将花费更少的时间来处理。这在较大的图上特别有用。我希望这对以后的人有所帮助。