On Wed, Nov 19, 2014 at 01:53:02PM -0500, David Feuer wrote: > I meant O(m log (mn)) = O(m (log m + log n)), because there are m appends, > building up from O(n) to O(mn), but it really doesn't matter because we can > easily do better. Indeed it's moot, but appending a tree of size n to one of size mn costs O(log n).