# [Haskell-cafe] Splay tree in a pattern matching way

Xinyu LIU liuxinyu95 at gmail.com
Sat Oct 23 23:06:07 EDT 2010

```Hi,

I checked the Hackage Splay Tree implementation from.

Basically it's based on the strategy mentioned in Okasaki's ``Purely
Functional Programming'', Chapter 5.4.
That in case we traverse twice in left (or right), we rotate the tree, so in
long term of view, the tree turns more and more balanced.

There are 3 cases for splay: zig-zig case, zig-zag case and zig case
according to
http://en.wikipedia.org/wiki/Splay_tree

So I wonder if Splay tree can also be implemented by using pattern matching
in a similar way as red black tree.

Here is a draft source code:

>>>
data STree a = E -- Empty
| Node (STree a) a (STree a) -- left, element, right
deriving (Eq, Show)

-- splay by pattern matching
splay :: (Eq a) => STree a -> a ->STree a
-- zig-zig
splay t@(Node (Node (Node a x b) p c) g d) y =
if x == y then Node a x (Node b p (Node c g d)) else t
splay t@(Node a g (Node b p (Node c x d))) y =
if x == y then Node (Node (Node a g b) p c) x d else t
-- zig-zag
splay t@(Node (Node a p (Node b x c)) g d) y =
if x == y then Node (Node a p b) x (Node c g d) else t
splay t@(Node a g (Node (Node b x c) p d)) y =
if x == y then Node (Node a g b) x (Node c p d) else t
-- zig
splay t@(Node (Node a x b) p c) y = if x == y then Node a x (Node b p
c) else t
splay t@(Node a p (Node b x c)) y = if x == y then Node (Node a p b) x
c else t
-- otherwise
splay t _ = t

-- insert by using pattern matching
insert :: (Ord a) => STree a -> a -> STree a
insert E y = Node E y E
insert (Node l x r) y
| x < y     = splay (Node (insert l y) x r) y
| otherwise = splay (Node l x (insert r y)) y
<<<

I am not quite sure if the solution is correct and what about the
efficiency of this code.

Regards.

--
Larry. LIU Xinyu