Python Leetcode 752：用 BFS 开锁 TLE

我试图用BFS解决leetcode问题https://leetcode.com/problems/open-the-lock/，并提出了以下方法.

def openLock(self, deadends: List[str], target: str): -> int
      def getNeighbors(node) ->List[str]:
            neighbors = []
            dirs = [-1, 1]
            for i in range(len(node)):
                for direction in dirs:
                    x = (int(node[i]) + direction) % 10
                    neighbors.append(node[:i] + str(x) + node[i+1:])
            return neighbors
                        
        start = '0000'
        if start in deadends:
            return -1
        queue = deque()
        queue.append((start, 0))
        turns = 0
        visited = set()
        deadends = set(deadends)
        while queue:
            state, turns = queue.popleft()
            visited.add(state)
            if state == target:
                return turns
            for nb in getNeighbors(state):
                if nb not in visited and nb not in deadends:   
                    queue.append((nb, turns+1))
        return -1

然而，这段代码遇到了一个超过时间限制的异常，它只通过了测试用例的2/48.例如，在一个测试用例中，死区是["8887","8889","8878","8898","8788","8988","7888","9888"]，目标是"8888"(开始总是"0000")，我的解决方案是TLE.我注意到，如果我本质上更改了一行并将相邻 node 添加到内部循环中的访问集，那么我的解决方案将通过该测试用例和所有其他测试用例.

for nb in getNeighbors(state):
     if nb not in visited and nb not in deadends:   
         visited.add(nb)
         queue.append((nb, turns+1))

然后，我的while循环变为

while queue:
            state, turns = queue.popleft()
            if state == target:
                return turns
            for nb in getNeighbors(state):
                if nb not in visited and nb not in deadends:   
                    visited.add(nb)
                    queue.append((nb, turns+1))

有人能解释为什么第一种方法会遇到TLE，而第二种方法不会？