The rose trees implemented in Data.Tree behave in the same way as the applicative style on lists (ie all possible combinations/permutations). I need a different behavior: one-to-one application. So, using the Data.Tree code, if I start with this tree: *Main Data.Traversable> putStrLn $ drawTree (fmap show temp2) 1 | +- 2 | `- 3 and then try this: *Main Data.Traversable> putStrLn $ drawTree (fmap show $ sequenceA [temp2,temp2]) [1,1] | +- [1,2] | +- [1,3] | +- [2,1] | | | +- [2,2] | | | `- [2,3] | `- [3,1] | +- [3,2] | `- [3,3] It gives all the combinations/permutations. I need this: *Main Data.Traversable> putStrLn $ drawTree (fmap show $ seqA [temp2,temp2]) [1,1] | +- [2,2] | `- [3,3] But then only way I could get that result was to (re-)write my own sequenceA function (called it seqA), which I suspect shouldn't be necessary. If I use the built-in sequenceA, I get (only) this: *Main Data.Traversable> putStrLn $ drawTree (fmap show $ sequenceA [temp2,temp2]) [1,1] Here's my code: data Tree a = Node {fromNode :: !a, fromNodeForest :: !(Forest a)} deriving (Show, Eq) type Forest a = [Tree a] zipWithForest = zipWith emptyForest = [] instance Functor Tree where fmap f (Node x subtrees) = Node (f x) (fmap (fmap f) subtrees) instance Applicative Tree where pure x = Node x emptyForest (<*>) (Node f tfs) (Node cargo subtrees) = Node (f cargo) (zipWithForest (<*>) tfs subtrees) seqA :: (Applicative f) => [f a] -> f [a] seqA (x:xs) = foldr (liftA2 (:)) (fmap (const []) x) (x:xs) I think I've got something wrong in how I've defined the Applicative Functor methods, and "cheating" by writing a custom sequenceA function to work around the deficiency. What is the proper fix? thanks again, travis