Question

我正在尝试实现DFS算法，以确定start节点和target节点之间是否存在路径。这是我到目前为止的代码：

# Depth-first search
def find_path2(s, t):
    s.visited = True

    if s.data == t.data:
        return True

    for node in s.neighbors:
        if not node.visited:
            return find_path2(graph, node, t)


node_0 = Node(0)
node_1 = Node(1)
node_2 = Node(2)
node_3 = Node(3)
node_4 = Node(4)
node_5 = Node(5)
node_6 = Node(6)

node_0.neighbors = [node_1]
node_1.neighbors = [node_2]
node_2.neighbors = [node_3, node_0]
node_3.neighbors = [node_2]
node_4.neighbros = [node_6]
node_5.neighbros = [node_4]
node_6.neighbors = [node_5]

start = node_2
target = node_0


if find_path2(start, target):
    print("There is a path between {} and {}".format(start.data, target.data))
else:
    print("There is no path between {} and {}".format(start.data, target.data))

node_2将node_3和node_0都作为邻居，因此应打印出它们之间存在路径。我理解return语句在第一次执行期间退出for循环，因为return语句退出函数，因此从不访问node_0。

我的问题是，最优雅的方式是什么？谢谢！

Answer 1

如果找到了您正在寻找的节点，您需要确保只从邻居的循环返回：

def find_path2(s, t):
    s.visited = True

    if s.data == t.data:
        return True

    for node in s.neighbors:
        if not node.visited:
            if find_path2(node, t):
                return True
    return False

深度优先搜索，递归，For循环和返回

1 个答案: