tree - Level-order in Haskell -

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i have structure tree , want print tree levels.

data tree = nd [tree a] deriving show type nd = string tree = nd "a" [nd "b" [nd "c" [],                        nd "g" [nd "h" [],                                nd "i" [],                                nd "j" [],                                nd "k" []]],                nd "d" [nd "f" []],                nd "e" [nd "l" [nd "n" [nd "o" []]],                        nd "m" []]] preorder (nd x ts) = x : concatmap preorder ts postorder (nd x ts) = (concatmap postorder ts) ++ [x]

but how levels? "levels tree" should print ["a", "bde", "cgflm", "hijkn", "o"]. think "iterate" suitable function purpose, cannot come solution how use it. me, please?

you need compute levels of subtrees , zip them after root:

levels :: tree [a] -> [[a]] levels (nd bs) = : foldr (zipwith' (++)) [] (map levels bs)

sadly, zipwith doesn't right thing, can instead use:

zipwith' f xs [] = xs zipwith' f [] xs = xs zipwith' f (x:xs) (y:ys) = f x y : zipwith' f xs ys

update: there concern (which agree with) asked little weird it's not generic breadth-first tree list convertor. want concat result of:

levels' (nd bs) = [a] : foldr (zipwith' (++)) [] (map levels' bs)

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