07.28.07

Exercise 7.1

Posted in Uncategorized at 9:37 pm by admin

data Tree a = Leaf a
            | Branch (Tree a) (Tree a)
              deriving Show
 
foldTree :: (a -> b -> b) -> (b -> b -> b) -> b -> Tree a -> b
foldTree fLeaf _ init (Leaf x) = fLeaf x init
foldTree fLeaf fBranch init (Branch x y) = fBranch x' y'
    where x' = foldTree fLeaf fBranch init x
          y' = foldTree fLeaf fBranch init y
 
fringe :: Tree a -> [a]
fringe t = foldTree (:) (++) [] t
 
treeSize :: Tree a -> Int
treeSize t = foldTree (\x y -> 1 + y) (+) 0 t
 
treeHeight :: Tree a -> Int
treeHeight t = foldTree (\x y -> 0) (\x y -> 1 + max x y) 0 t

The key to a tree fold was realising that two functions were needed: one for leaves and one for branches. In general, I think any fold would require a function per constructor for the data structure it works on.

2 Comments »

  1. Wieslaw Poszewiecki said,

    April 14, 2008 at 1:59 am

    Following also seems to work:

    foldTree _ init (Leaf x) = init x
    foldTree op init (Branch t1 t2) = (foldTree op init t1) `op` (foldTree op init t2)

    fringe’ = foldTree (++) (:[])
    treeSize’= foldTree (+) (const 1)
    treeHeight’= foldTree (\x y->max x y+1) (const 0)

  2. Ferran Maylinch said,

    April 7, 2011 at 1:45 pm

    Yes, I also think that there is no need for an “init” value.
    You only need a leaf function.

    My solution is like Wieslav’s but for beginners 🙂

    foldt fl fb (Leaf x) = fl x
    foldt fl fb (Branch t1 t2) = fb t1′ t2′
    where
    t1′ = foldt fl fb t1
    t2′ = foldt fl fb t2

    fringe t = foldt tolist (++) t where tolist x = [x]
    treeSize t = foldt one (+) t where one x = 1
    treeHeight t = foldt zero max1 t
    where
    zero x = 0
    max1 t1 t2 = 1 + max t1 t2

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