# 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)
```