Find Element in Tree Given a List of Directions

Looking at Learn You a Haskell, I implemented a function that, given a binary tree, i.e. left and right branches, give a list of Directions and give the element at that location.

Here's the Tree:

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

And then the functions:

elemAt :: Directions -> Tree a -> a  
elemAt []     (Node x _ _) = x
elemAt (R:ds) (Node _ _ r) = elemAt ds r
elemAt (L:ds) (Node _ l _) = elemAt ds l

But, I re-wrote it to be safe (maybe not)

maybeElemAt :: Directions -> Tree a -> Maybe a  
maybeElemAt []     (Node x _ _)     = Just x
maybeElemAt (R:ds) (Node _ _ Empty) = Nothing
maybeElemAt (R:ds) (Node _ _ r)     = maybeElemAt ds r
maybeElemAt (L:ds) (Node _ Empty _) = Nothing
maybeElemAt (L:ds) (Node _ l _)     = maybeElemAt ds l

Testing

anotherTree :: Tree Char
anotherTree = Node 'A' (Node 'B' Empty Empty) Empty

ghci> maybeElemAt [R,R,R] anotherTree
Nothing
ghci> maybeElemAt [R] anotherTree
Nothing
ghci> maybeElemAt [L] anotherTree
Just 'B'

Please critique my implementations of either function.