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237 changes: 237 additions & 0 deletions data_structures/binary_tree/splay_tree.py
Original file line number Diff line number Diff line change
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"""
Splay Tree - a self-adjusting binary search tree.

A splay tree is a binary search tree with the additional property that
recently accessed elements are quick to access again. Every access (search,
insert or delete) moves the target node to the root through a sequence of
rotations called "splaying". This gives an amortized time complexity of
O(log n) per operation and makes the tree very efficient when the access
pattern has locality of reference (a small subset of keys is touched often).

Reference: https://en.wikipedia.org/wiki/Splay_tree
"""

from __future__ import annotations

from collections.abc import Iterator
from dataclasses import dataclass, field


@dataclass
class Node:

@cclauss cclauss Sep 10, 2026

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Please simplify by making Node a dataclass.

"""
A single node of a splay tree.

The ``left`` and ``right`` children are excluded from ``repr`` so that a
node prints compactly instead of recursively dumping the whole subtree.

>>> Node(10)
Node(key=10)
"""

key: int
left: Node | None = field(default=None, repr=False)
right: Node | None = field(default=None, repr=False)


class SplayTree:
"""
A self-adjusting binary search tree.

>>> tree = SplayTree()
>>> tree.insert(10)
>>> tree.insert(20)
>>> tree.insert(30)
>>> tree.root.key # last inserted key is splayed to the root
30
>>> tree.search(10)
True
>>> tree.root.key # the searched key is now the root
10
>>> tree.search(99)
False
>>> list(tree)
[10, 20, 30]
"""

def __init__(self) -> None:
self.root: Node | None = None

def _rotate_right(self, node: Node) -> Node:
"""
Perform a right rotation around ``node`` and return the new subtree root.

node left
/ \\ / \\
left c --> a node
/ \\ / \\
a b b c
"""
left = node.left
assert left is not None
node.left = left.right
left.right = node
return left

def _rotate_left(self, node: Node) -> Node:
"""
Perform a left rotation around ``node`` and return the new subtree root.

node right
/ \\ / \\
a right --> node c
/ \\ / \\
b c a b
"""
right = node.right
assert right is not None
node.right = right.left
right.left = node
return right

def _splay(self, root: Node | None, key: int) -> Node | None:
"""
Splay the node with ``key`` (or the last node on the search path if
``key`` is absent) to the root of the subtree and return the new root.
This uses the classic bottom-up recursive formulation.
"""
if root is None or root.key == key:
return root

if key < root.key:
if root.left is None:
return root
if key < root.left.key:
# Zig-Zig (left left)
root.left.left = self._splay(root.left.left, key)
root = self._rotate_right(root)
elif key > root.left.key:
# Zig-Zag (left right)
root.left.right = self._splay(root.left.right, key)
if root.left.right is not None:
root.left = self._rotate_left(root.left)
return root if root.left is None else self._rotate_right(root)
else:
if root.right is None:
return root
if key > root.right.key:
# Zig-Zig (right right)
root.right.right = self._splay(root.right.right, key)
root = self._rotate_left(root)
elif key < root.right.key:
# Zig-Zag (right left)
root.right.left = self._splay(root.right.left, key)
if root.right.left is not None:
root.right = self._rotate_right(root.right)
return root if root.right is None else self._rotate_left(root)

def insert(self, key: int) -> None:
"""
Insert ``key`` into the tree and splay it to the root.

>>> tree = SplayTree()
>>> for key in (5, 3, 8, 3): # duplicate keys are ignored
... tree.insert(key)
>>> list(tree)
[3, 5, 8]
>>> tree.root.key # the duplicate access splays 3 back to the root
3
"""
if self.root is None:
self.root = Node(key)
return

self.root = self._splay(self.root, key)
assert self.root is not None
if self.root.key == key:
return # key already present, it is now at the root

node = Node(key)
if key < self.root.key:
node.right = self.root
node.left = self.root.left
self.root.left = None
else:
node.left = self.root
node.right = self.root.right
self.root.right = None
self.root = node

def search(self, key: int) -> bool:
"""
Return whether ``key`` is present and splay the last accessed node.

>>> tree = SplayTree()
>>> tree.search(1)
False
>>> for key in (40, 20, 60):
... tree.insert(key)
>>> tree.search(20)
True
>>> tree.root.key
20
"""
self.root = self._splay(self.root, key)
return self.root is not None and self.root.key == key

def delete(self, key: int) -> None:
"""
Remove ``key`` from the tree if it is present.

>>> tree = SplayTree()
>>> for key in (10, 20, 30, 40):
... tree.insert(key)
>>> tree.delete(20)
>>> list(tree)
[10, 30, 40]
>>> tree.delete(99) # deleting an absent key is a no-op
>>> list(tree)
[10, 30, 40]
>>> for key in (10, 30, 40):
... tree.delete(key)
>>> list(tree)
[]
"""
if self.root is None:
return

self.root = self._splay(self.root, key)
assert self.root is not None
if self.root.key != key:
return # key not found

left, right = self.root.left, self.root.right
if left is None:
self.root = right
else:
# Splay the maximum of the left subtree to its root; it has no
# right child, so the right subtree can be attached there.
left = self._splay(left, key)
assert left is not None
left.right = right
self.root = left

def __iter__(self) -> Iterator[int]:
"""
Yield the keys of the tree in ascending (in-order) order.

>>> tree = SplayTree()
>>> for key in (7, 2, 9, 4, 1):
... tree.insert(key)
>>> list(tree)
[1, 2, 4, 7, 9]
"""

def in_order(node: Node | None) -> Iterator[int]:
if node is not None:
yield from in_order(node.left)
yield node.key
yield from in_order(node.right)

yield from in_order(self.root)


if __name__ == "__main__":
import doctest

doctest.testmod()
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