Most concise way to fold tree structures

Haskell, 37 35:

data T a=a:*[T a]

(f%g)z(x:*t)=x`f`foldr(g.(f%g)z)z t

not counting the type definition. With it, 54 52 (shortened by using infix operator, similarly to the answer by proud haskeller, but in a different way).

Ungolfed:

data Tree a = Node a [Tree a]

foldTree :: (a -> b -> c) -> (c -> b -> b) -> b -> Tree a -> c
foldTree f g z (Node x t) = f x $ foldr (g . foldTree f g z) z t
                       -- = f x . foldr g z . map (foldTree f g z) $ t

--      1
--     / \
--    2   3
--   / \   \
--  4   5   6

t = Node 1 [Node 2 [Node 4 [], Node 5 []],
            Node 3 [Node 6 []]]

result = foldTree (+) (*) 1 t   -- ((+)%(*)) 1 t        {-
       = 1 + product [2 + product[ 4 + product[], 5 + product[]],
                      3 + product[ 6 + product[]]]
       = 1 + product [2 + 5*6, 3 + 7]
       = 321                                            -}

   -- product [a,b,c,...,n] = foldr (*) 1 [a,b,c,...,n] 
   --     = a * (b * (c * ... (n * 1) ... ))             

This is the redtree ("reduce tree") function from John Hughes paper, "Why Functional Programming Matters", where it is presented in a more verbose formulation.

F#, 70 61

let rec r f g a (Node(n,c))=f n<|List.foldBack(g<<r f g a)c a

Not going to win the contest, but I think you can't get less with F#

Haskell, 35

data T a=a:>[T a]
h(%)g a(e:>l)=e%foldr(g.h(%)g a)a l

the data type declaration is not counted.