BFS uses a FIFO queue: the first node enqueued is the first one dequeued. The crucial detail is marking a node visited when you enqueue it, not when you pop it; otherwise the same node can be enqueued many times by different neighbors. collections.deque gives O(1) popleft; using a plain list with pop(0) accidentally turns the algorithm into O(n^2) and is a top interview gotcha. The same loop body works for any iterable graph: dictionaries, adjacency matrices, or even on-the-fly neighbor functions.

Distances fall out of BFS for free: the first time you see a node it must be at the smallest possible edge count (BFS visits in order of distance), so set dist[nbr] = dist[node] + 1 at the moment you enqueue it. Nodes that never appear in the dict are unreachable from the source. This is the right algorithm for any 'fewest moves' question on an unweighted graph: chess knight tours, word ladders, smallest number of bus transfers. For weighted graphs you need Dijkstra instead.

When the algorithm cares about levels (binary tree level order, 'snapshot of frontier per round'), keep a list of the current level instead of a single queue. After processing every node in level, build next_level from their unvisited neighbors and yield the snapshot. This avoids the bookkeeping of attaching a (node, depth) tuple to every queue entry and naturally separates rounds. Use it for animated traversals, k-th-level queries, and any 'closest things first, ties broken arbitrarily' problem.