a)
From my understanding, yes, the lists produced are always going to be equal?
given the same tree

b)
No, they don't, remember than when we `append` two lists, the cons cells for
the first list have to all be reconstructed, which is an additional O(n)
steps, where n is len(first-list).

This means that when we break up a tree in (tree->list-1 tree) T(n) = T(n/2) +
O(n) + T(n/2)

tree->list-2 has one cons operation, which is constant, for each entry, which
means the order of growth is O(n)

On the other hand, tree->list-1 has order of growth: O(n log n)