FP Laboratory 10

Binary Trees

• Create a data type Tree that defines binary tree where values are stored in leaves and also in branches.

data Tree a = Leaf a 
            | Branch a (Tree a) (Tree a) deriving (Show)

• Prepare an example of a binary tree.

testTree1 :: Tree Int            
testTree1 = Branch 12 (Branch 23 (Leaf 34) (Leaf 45)) (Leaf 55)

testTree2 :: Tree Char            
testTree2 = Branch 'a' (Branch 'b' (Leaf 'c') (Leaf 'd')) (Leaf 'e')

• Create a function that sums all values stored in the tree.

sum' :: Tree Int -> Int

sum' :: Tree Int -> Int
sum' (Leaf x) = x
sum' (Branch x l r) = sum' l + x + sum' r

• Create a function that extracts all values from the tree into an list.

toList :: Tree a -> [a]

toList :: Tree a -> [a]
toList (Leaf x) = [x]
toList (Branch x l r) = toList l ++ [x] ++ toList r

• Create a function that finds a maximum value stored in the tree.

maxTree :: Ord a => Tree a -> a

• Create a function that returns the depth of a tree.

depthTree :: Tree a -> Int

• Create a function that returns as list elements bigger than a given value.

getGreaterElements :: Ord a => Tree a -> a -> [a]

• One possibility how to represent a tree in a textual form is a(b(d,e),c(e,f(g,h))). Create functions that are able to read and store a tree in such a notation.

toString :: Show a => Tree a -> String
fromString :: Read a => String -> Tree a

toString :: Show a => Tree a -> String
toString (Leaf x) = show x
toString (Branch x l r) = show x ++ "(" ++ (toString l) ++ "," ++ (toString r) ++ ")"

fromString :: Read a => String -> Tree a
fromString inp = fst (fromString' inp) where
  fromString' :: Read a => String -> (Tree a,String)
  fromString' inp = 
    let
      before = takeWhile (\x ->  x /='(' &&  x /=',' &&  x/=')') inp 
      after = dropWhile (\x ->  x /='(' &&  x /=',' && x/=')') inp
      value = read before
    in if null after || head after /= '(' then (Leaf value, after) else 
        let
          (l,after') = fromString' (tail after)
          (r,after'') = fromString' (tail after') 
        in (Branch value l r, tail after'')

Additional exercises

• Create a function that counts all leaves in the tree.

leafCount :: Tree a -> Int

• Create a function that counts all branches in the tree.

branchCount :: Tree a -> Int

• Create a function that checks whether a given element is stored in the tree.

contains :: Eq a => Tree a -> a -> Bool

• Create a function that returns a number of elements greater than a given value.

greaterThan :: Ord a => Tree a -> a -> Int

• Consider the following alternative definition of the binary tree.

data Tree2 a = Null | Branch a (Tree2 a) (Tree2 a)

• Is this definition equivalent to the previous one? If not, explain why and give an example of a tree that can be constructed with the first definition but not with the second one or vice versa.

• Implement all functions above adjusted for the alternative definition of the binary tree.

Abstract data types

• Create an implementation of the abstract data type Stack with following functions:

push :: a -> Stack a -> Stack a
pop :: Stack a -> Stack a
top :: Stack a -> a
isEmpty :: Stack a ->Bool

module Stack(Stack, emptyS, push, pop, top, isEmpty) where
  data Stack a = Stack [a] deriving Show

  emptyS :: Stack a
  emptyS = Stack []

  push :: a -> Stack a -> Stack a
  push x (Stack y) = Stack (x:y)
  
  pop :: Stack a -> Stack a
  pop (Stack (_:xs)) = Stack xs

  top :: Stack a -> a
  top (Stack (x:_)) = x

  isEmpty :: Stack a ->Bool
  isEmpty (Stack []) = True
  isEmpty _ = False

• Create an implementation of the abstract data type Queue with following functions:

isEmpty :: Queue a -> Bool
addQ :: a -> Queue a -> Queue a
remQ :: Queue a -> (a, Queue a)

module Queue(Queue, emptyQ, isEmptyQ, addQ, remQ) where
    data Queue a = Qu [a] deriving Show

    emptyQ :: Queue a
    emptyQ = Qu []
    
    isEmptyQ :: Queue a -> Bool
    isEmptyQ (Qu q) = null q
    
    addQ :: a -> Queue a -> Queue a
    addQ x (Qu xs) = Qu (xs++[x])
    
    remQ :: Queue a -> (a,Queue a)
    remQ q@(Qu xs) | not (isEmptyQ q) = (head xs, Qu (tail xs))
                   | otherwise        = error "remQ"