https://www.youtube.com/watch?v=46dksIrx6jQ&feature=youtu.be

solution.hs

rev :: [a] -> [a]
rev = foldl (\acc x -> x : acc) []
--rev = foldl (flip (:)) []

prefixes :: [a] -> [[a]]
prefixes = foldr (\x acc -> [x] : (map ((:) x) acc)) []

lagrange :: [(Float, Float)] -> Float -> Float
lagrange xs x = foldl (\acc (xj,y) -> acc + (y * l xj)) 0 xs
 where
  l xj = foldl (
       \acc (xk,_) -> 
	       if xj==xk then 
		     acc 
		   else 
		     acc * ((x-xk)/(xj-xk))
		) 1 xs
		
		
data Trie a = Leaf a | Node a [Trie a]

foldtrie :: (b -> a -> b) -> b -> Trie a -> b
foldtrie f acc (Leaf x) = f acc x
foldtrie f acc (Node x xs) = foldl f' (f acc x) xs
 where
  f' acc t = foldtrie f acc t

t = Node 'c' [(Node 'a' [Leaf 'r', Leaf 't']),(Node 'o' [Node 'o' [Leaf 'l']])]

The output is dependent on the function that is used with foldtrie. It is equivalent with foldl for lists where the leftmost leafs of the trie are the first ones that the function is used on. So if we want to create a string representing the leafs of the tree (going from left to right) we have to append the elements to the accumulator: foldtrie (\acc x -> acc ++ [x]) "" t