Challenge 2: A Functor Instance for Tree a — Possible Solution
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Main.hs:

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

instance Functor Tree where
  fmap _ Leaf = Leaf
  fmap f (Node l v r) = Node (fmap f l) (f v) (fmap f r)

sampleTree :: Tree Int
sampleTree = Node (Node Leaf 1 Leaf) 2 (Node Leaf 3 Leaf)

main :: IO ()
main = print (fmap (*10) sampleTree)

Output:

    Node (Node Leaf 10 Leaf) 20 (Node Leaf 30 Leaf)

Explanation:
The Functor instance mirrors Tree's own recursive shape exactly: Leaf
maps to Leaf (nothing to transform in an empty subtree), and a Node
recursively fmaps over both its left and right subtrees while
applying f directly to its own value. Calling fmap (*10) on
sampleTree multiplies every value in the tree by 10 -- 1, 2, and 3
become 10, 20, and 30 -- while the tree's overall shape (which nodes
are Leaf vs Node, and where) stays completely unchanged.

WHY THIS WORKS AS AN ANSWER
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This writes the exact Functor instance the chapter introduces for
Tree a, applies it to a real multi-node tree (not just a trivial
single-node example), and confirms both the transformed values and
the preserved tree structure in the output.