99 questions/Solutions/67A

A string representation of binary trees

Somebody represents binary trees as strings of the following type:

a(b(d,e),c(,f(g,)))

a) Write a Prolog predicate which generates this string representation, if the tree is given as usual (as nil or t(X,L,R) term). Then write a predicate which does this inverse; i.e. given the string representation, construct the tree in the usual form. Finally, combine the two predicates in a single predicate tree_string/2 which can be used in both directions.

treeToString :: Tree Char -> String
treeToString Empty = ""
treeToString (Branch x Empty Empty) = [x]
treeToString (Branch x l r) =
        x : '(' : treeToString l ++ "," ++ treeToString r ++ ")"

stringToTree :: (Monad m) => String -> m (Tree Char)
stringToTree "" = return Empty
stringToTree [x] = return $ Branch x Empty Empty
stringToTree str = tfs str >>= \ ("", t) -> return t
    where tfs a@(x:xs) | x == ',' || x == ')' = return (a, Empty)
          tfs (x:y:xs)
                | y == ',' || y == ')' = return (y:xs, Branch x Empty Empty)
                | y == '(' = do (',':xs', l) <- tfs xs
                                (')':xs'', r) <- tfs xs'
                                return $ (xs'', Branch x l r)
          tfs _ = fail "bad parse"

Note that the function stringToTree works in any Monad.

The following solution for 'stringToTree' uses Parsec:

import Text.Parsec.String
import Text.Parsec hiding (Empty)


pTree :: Parser (Tree Char)
pTree = do
    pBranch <|> pEmpty

pBranch = do
    a <- letter
    char '('
    t0 <- pTree 
    char ','
    t1 <- pTree
    char ')'
    return $ Branch a t0 t1

pEmpty =
    return Empty

stringToTree str =
    case parse pTree "" str of
        Right t -> t
        Left e  -> error (show e)