Leetcode 1971. Find if Path Exists in Graph [Python]

BFS 从start遍历到end,每一次que弹出节点是end,返回true,否则,把此节点加入到seen set中,并入队。遍历完成后,未找到end节点,代表和start直接或间接相连的节点中没有end节点。返回false。注意特殊情况,只有一个节点时。

class Solution:
    def validPath(self, n: int, edges: List[List[int]], start: int, end: int) -> bool:
        if n == 1:return True
        dic = collections.defaultdict(set)
        for n,m in edges:
            dic[n].add(m)
            dic[m].add(n)
        que = collections.deque()
        que.append(start)
        seen = set()
        while que:
            size = len(que)
            for _ in range(size):
                curnode = que.popleft()
                for nextnode in dic[curnode]:
                    if nextnode == end:
                        return True
                    else:
                        if nextnode not in seen:
                            seen.add(nextnode)
                            que.append(nextnode)
        return False
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