This comment seems a little misleading --- the reason for prepending the visitableNeighbors is because we're treating pending as a stack rather than a queue, not because it is more efficient (though I agree it will be). Regardless of which list is shorter, prepending the visitableNeighbors will take constant time per new neighbor, and hence $O(n)$ overall since each vertex occurs as a new neighbor at most once. Whereas appending visitableNeighbors will take time linear in the number of vertices in the queue, but the same vertices might be in the queue for multiple iterations, so we can end up doing more than $O(n)$ work appending.

the reason for prepending the visitableNeighbors is because we're treating pending as a stack rather than a queue, not because it is more efficient (though I agree it will be)

Well, I originally went about this intending to implement BFS, until I got to this line and realized that appending to the pending list would be expensive (and I couldn't be bothered to use Seq instead of []). So I decided to prepend instead, which turned it into DFS.