Johan Tibell <johan.tibell@gmail.com> writes:

> It's not true that
>
>     fold f z == foldr f z . elems
>
> in general. It only holds if `z` is an identity for `f` as `z` is used
> at every leaf in the tree.

Hey Johan,

   -- | /O(n)/. Post-order fold.
   foldr :: (k -> a -> b -> b) -> b -> Map k a -> b
   foldr _ z Tip              = z
   foldr f z (Bin _ kx x l r) = foldr f (f kx x (foldr f z r)) l

Note the third parameter to the recursive foldr call -- the "z" you see
at the Tips is not the same "z" that was originally passed in.