Mark it when adding it. Otherwise several parents can queue the same vertex before its first removal. First discovery already gives the shortest edge-count distance under BFS’s layer ordering.

from collections import deque

def distances(graph, start):
    distance = {start: 0}
    queue = deque([start])
    while queue:
        node = queue.popleft()
        for neighbor in graph.get(node, []):
            if neighbor not in distance:
                distance[neighbor] = distance[node] + 1
                queue.append(neighbor)
    return distance

graph = {"A": ["B", "C"], "B": ["D"], "C": ["D"], "D": []}
assert distances(graph, "A") == {"A": 0, "B": 1, "C": 1, "D": 2}

After processing A, the queue holds B, C, both at distance 1. Processing B appends D at distance 2 behind the remaining distance-1 work. No longer route can discover D before every possible shorter predecessor has had its turn.

With adjacency lists, BFS takes O(V+E) time over the reachable portion and O(V) extra space in the worst case. Use deque.popleft(); removing index zero from a Python list shifts elements.

To reconstruct one shortest path, store a parent on first discovery and follow parents backward from the destination. To count shortest paths, also accumulate counts from other parents at the preceding distance. “Already discovered” does not mean every question should ignore every later edge.

Analogy limit: a real ripple over unequal terrain is not this algorithm. BFS minimizes edge count; arbitrary edge weights require another argument.

The direct one-edge route, even though its cost is worse. For nonnegative varying weights, use a cost-aware method such as Dijkstra.