What constitutes a fold for types other than list?

Question

Consider a single-linked list. It looks something like

data List x = Node x (List x) | End

It is natural to define a folding function such

Answer 1

A fold replaces every constructor with a function.

For example, foldr cons nil replaces every (:) with cons and [] with nil:

foldr cons nil ((:) 1 ((:) 2 [])) = cons 1 (cons 2 nil)

For a tree, foldTree branch leaf replaces every Branch with branch and every Leaf with leaf:

foldTree branch leaf (Branch (Branch (Leaf 1) (Leaf 2)) (Leaf 3))
    = branch (branch (leaf 1) (leaf 2)) (leaf 2)

This is why every fold accepts arguments that have the exact same type as the constructors:

foldr :: (a -> list -> list) -> list -> [a] -> list

foldTree :: (tree -> tree -> tree) -> (a -> tree) -> Tree a -> tree