## Exercise 7.1

```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 Responses to “Exercise 7.1”

1. Wieslaw Poszewiecki says:

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 says:

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