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Question:

For my Algorithms & Data Structures class, I've been tasked with implementing a splay tree in Haskell. My algorithm for the splay operation is as follows:

- If the node to be splayed is the root, the unaltered tree is returned.
- If the node to be splayed is one level from the root, a zig operation is performed and the resulting tree is returned.
- If the node to be splayed is two or more levels from the root, a zig-zig or zig-zag operation is performed on the result of splaying the subtree starting at that node, and the resulting tree is returned.

This is valid according to my teacher. However, the Wikipedia description of a splay tree says the zig step "will be done only as the last step in a splay operation" whereas in my algorithm it is the first step in a splay operation.

I want to implement a splay tree that performs the zig operation last instead of first, but I'm not sure how it would best be done. It seems to me that such an algorithm would become more complex, seeing as how one needs to find the node to be splayed before it can be determined whether a zig operation should be performed or not.

How can I implement this in Haskell (or some other functional language)?

## Example

In this example we search for the value 4, prompting us to splay it to the top of the tree.

### My algorithm (zig as the first step)

1 1 4 \ \ / 2 zig 2 zig-zig 2 \ --> \ ------> / \ 3 4 1 3 \ / 4 3

### Wikipedia algorithm (zig as the last step)

1 1 4 \ \ / 2 zig-zig 4 zig 1 \ ------> / --> \ 3 3 3 \ / / 4 2 2

Both trees are valid, but they have different structures. I want to implement the second one in a functional language, preferably Haskell.

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Solution:1

The key is to build a path to the value to be splayed, then rebuild the tree from the bottom, two levels at a time if possible (so that the zig-zip vs. zig-zag determination can be made):

`data Tree a = Empty | Node a (Tree a) (Tree a) deriving (Eq, Show) data Direction = LH | RH deriving (Eq, Show) splay :: (Ord a) => a -> Tree a -> Tree a splay a t = rebuild $ path a t [(undefined,t)] where path a Empty ps = ps path a n@(Node b l r) ps = case compare a b of EQ -> ps LT -> path a l $ (LH, l) : ps GT -> path a r $ (RH, r) : ps rebuild :: (Ord a) => [(Direction,Tree a)] -> Tree a rebuild ((_,n):[]) = n rebuild ((LH,x):(_,p):[]) = zigL x p rebuild ((RH,x):(_,p):[]) = zigR x p rebuild ((LH,x):(LH,p):(z,g):ps) = rebuild $ (z, zigzigL x p g):ps rebuild ((RH,x):(RH,p):(z,g):ps) = rebuild $ (z, zigzigR x p g):ps rebuild ((RH,x):(LH,p):(z,g):ps) = rebuild $ (z, zigzagL x p g):ps rebuild ((LH,x):(RH,p):(z,g):ps) = rebuild $ (z, zigzagR x p g):ps zigL (Node x a b) (Node p _ c) = Node x a (Node p b c) zigR (Node x a b) (Node p c _) = Node x (Node p c a) b zigzigL (Node x a b) (Node p _ c) (Node g _ d) = Node x a (Node p b (Node g c d)) zigzigR (Node x a b) (Node p c _) (Node g d _) = Node x (Node p (Node g d c) a) b zigzagL (Node x b c) (Node p a _) (Node g _ d) = Node x (Node p a b) (Node g c d) zigzagR (Node x b c) (Node p _ a) (Node g d _) = Node x (Node g d b) (Node p c a) `

You can find this code, along with runnable unit tests and quick checks in my repo.

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Solution:2

Are you sure you're reading the Wikipedia description correctly? There are three kinds of steps: "zig", "zig-zig", and "zig-zag". The "zig" step is *defined* to be something that happens only when `x`

is a child of the root. Despite the names, the "zig-zig" and "zig-zag" steps don't have "zig" steps as a first component.

It sounds to me like your implementation follows the Wikipedia description in this respect.

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Solution:3

You can ref this course, which has a very good lecture note with code in OCaml for Splay tree.

**Note:If u also have question or solution just comment us below or mail us on toontricks1994@gmail.com**

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