1 We'll give you a hint, but it will require some extra thought.
3 The hint is a solution to this exercise taken from the source code that accompanies the Glasgow Haskell Compiler (underneath */Control/Monad/State/Strict.hs*).
5 We're not going to massage it in any way. If you want to make use of it, you'll have to figure out for yourself what's going on. This should be within your reach at this point. See our page on
6 [[translating between OCaml Scheme and Haskell]] for guidance. See our [[state monad tutorial]] for explanation of `get` and `put`.
8 Also, you'll notice that this solution targets trees with labels on their inner nodes, instead of on their leaves. It shouldn't be too hard to get a similar strategy to work for leaf-labeled trees.
11 data Tree a = Nil | Node a (Tree a) (Tree a) deriving (Show, Eq)
14 numberTree :: Eq a => Tree a -> State (Table a) (Tree Int)
15 numberTree Nil = return Nil
16 numberTree (Node x t1 t2)
17 = do num <- numberNode x
20 return (Node num nt1 nt2)
22 numberNode :: Eq a => a -> State (Table a) Int
25 (newTable, newPos) <- return (nNode x table)
28 nNode:: (Eq a) => a -> Table a -> (Table a, Int)
30 = case (findIndexInList (== x) table) of
31 Nothing -> (table ++ [x], length table)
33 findIndexInList :: (a -> Bool) -> [a] -> Maybe Int
34 findIndexInList = findIndexInListHelp 0
35 findIndexInListHelp _ _ [] = Nothing
36 findIndexInListHelp count f (h:t)
39 else findIndexInListHelp (count+1) f t
43 -- numTree applies numberTree with an initial state:
45 numTree :: (Eq a) => Tree a -> Tree Int
46 numTree t = evalState (numberTree t) []
48 testTree = Node "Zero" (Node "One" (Node "Two" Nil Nil) (Node "One" (Node "Zero" Nil Nil) Nil)) Nil
51 ~~> Node 0 (Node 1 (Node 2 Nil Nil) (Node 1 (Node 0 Nil Nil) Nil)) Nil