ramchandra's blog

By ramchandra, 3 years ago, In English

This tutorial is about the splay tree, a self-balancing binary search tree.

Key features:
  • $$$O(\log n)$$$ amortized operations.

  • $$$O(n)$$$ construction time complexity (you can simply build a binary search tree)

  • Memory-efficient: no additional space needed apart from the binary search tree.

The key idea behind the splay tree is the splay function, which rotates nodes so as to balance the tree. Since this consists only of tree rotations, this does not affect the binary search tree ordering.

Tree rotation

A tree rotation inverts the relationship between a parent and a child node, without affecting the in order traversal of the binary search tree. See the animation below.

Tree rotation

(Animation source: Wikipedia user Tar-Elessar, licensed under CC-BY-SA 4.0)

Splay operation

During a splay step, we repeatedly do this until the node reaches the root: If there is a parent: Rotate the parent if the node's parent is on the same as the node (e.g. the node == node->parent->parent->left->left or node == node->parent->parent->right->right). Otherwise rotate the node. Then rotate the node.

This splay operation tends to balance the tree. The standard BBST operations can be implemented based on splay.

Implementing other operations

To insert a node, insert regularly as you would in a binary search tree (by walking down the binary search tree), and then splay the newly created node. Then set the root node to the newly created node.

To join two splay trees a and b, where all the keys in a are less than all the keys in b, splay the maximum element of a. Since the maximum element of a is now at the root, it has no right child (since it is the maximum element). So, simply attach b as the right child of a. a will then be the joined tree.

To erase a node, splay the node and then join the node's two child subtrees.

Implementation details

Note: you may not want to use recursion in a splay tree as that could lead to stack overflow in the worst-case, where the tree could have depth $$$O(n)$$$. Recursion is alright on Codeforces since it has a large stack limit.

In this implementation, the side function tells us which side the node is relative to its parent. We store the left and right children in an array to simplify the code. That way we can avoid writing two functions rotate_left and rotate_right. Instead, we just need to write one function rotate that handles all the rotation cases. Also, the helper function attach attaches a child node to a parent node. The helper function attach is used.

For finding a node, be sure to splay the last non null node visited to ensure $$$O(\log n)$$$ complexity. (Thanks to Narut for pointing this out)


Generally, since they are both BBSTs, splay tree can be used in lieu of treap. So these [treap problems] can be solved with splay tree as well.

Another application is the link-cut tree, a data structure used to represent a rooted forest. Stay tuned for a blog on link-cut tree in a few days.

#include <bits/stdc++.h>
using namespace std;
using ll = long long int;
/** @brief Implements a splay tree*/
template <typename T> struct SplayTree {
	struct Node {
		Node(T value_arg): value{value_arg} {}
		T value{}; //!< Value associated with node
		array<Node *, 2> child{}; //!< Left and right children
		Node *parent{}; //!< Pointer to parent
		//Node *path_parent{};
		bool side() const {
			/*! Returns true if child is on the right, and false otherwise*/
			return parent->child[1] == this;
		void rotate() {
			/*! Rotate node x around its parent */
			const auto p = parent;
			const bool i = side();

			if (p->parent) {
				p->parent->attach(p->side(), this);
			} else {
				parent = nullptr;
			p->attach(i, child[!i]);
			attach(!i, p);
		void splay() {
			/*! Splay node x. x will become the root of the tree*/
			for(;parent;rotate()) {
				if (parent->parent) {
					(side() == parent->side() ? parent: this)->rotate();
		array<Node *, 2> split() {
			const auto right = child[1];
			if (right) {
				right->parent = nullptr;
			this->right = nullptr;
			return {this, right};
		void attach(bool side, Node *const new_) {
			/*! Attach node new_ as the node's side children*/
			if (new_) {
				new_->parent = this;
			child[side] = new_;
	/** @brief Splay tree node */
	/*! Splay tree iterator */
	struct iterator {
		using iterator_category = bidirectional_iterator_tag;
		using value_type = T;
		using pointer = T *;
		using reference = T &;
		using difference_type = ll;
		void operator--() { advance<false>(); }
		void operator++() { advance<true>(); }
		const T &operator*() { return node->value; }
		Node *node;
		iterator(Node *node_arg) : node(node_arg) {}
		bool operator==(const iterator oth) const {
			return node == oth.node;
		bool operator!=(const iterator oth) const {return !(*this == oth);}
		template <bool dir> void advance() {
			if (node->child[1]) {
				node = extremum<!dir>(node->child[1]);
			for (; node->parent && node->side() == dir; node = node->parent)
			node = node->parent;
	Node *root{}; /*!< Root node */
	ll size_{};   /*!< Size of the splay tree*/
	SplayTree() {}
	~SplayTree() { destroy(root); }
	static void destroy(Node *const node) {
		/*! Destroy the subtree node */
		if (!node) {
		for (Node *const child : node->child) {
		delete node;
	void insert(Node *const x) {
		/*! Insert node x into the splay tree*/
		if (!root) {
			root = x;
		auto y = root;
		while (true) {
			auto &nw = y->child[x->value > y->value];
			if (!nw) {
				nw = x;
				nw->parent = y;
				root = nw;
			y = nw;
	void insert(const T &key) {
		/*! Insert key key into the splay tree*/
		insert(new Node{key});
	void erase(Node *const x) {
		/*! Erase node x from the splay tree*/
		root = join(x->child[0], x->child[1]);
		delete x;
	/** @brief Erases the node with key key from the splay tree */
	void erase(const T &key) { erase(find(key)); }
	template <bool i> static Node *extremum(Node * x) {
		/*! Return the extremum of the subtree x. Minimum if i is false,
		 * maximum if i is true.*/
		for(; x->child[i]; x = x->child[i]);
		return x;
	static Node *join(Node *const a, Node *const b) {
		if (!a) {
			b->parent = nullptr;
			return b;
		Node *const mx = extremum<true>(a);
		assert(mx->child[1] == nullptr);
		mx->child[1] = b;
		mx->parent = nullptr;
		return mx;
    /*! Returns node with key key*/
	Node *find(const T &key) {
		auto x = root;
		while(x && key != x->value){
			const auto next = x->child[key > x->value];
			x = next;
		return x;
     * @brief Returns the number of nodes in the splay tree.
    size_t size() const { return size_; }
	bool empty() const { return size() == 0; }
	iterator begin() { return iterator{extremum<false>(root)}; }
	iterator end() { return iterator{nullptr}; }
int main() {
	/*! Tests the splay tree*/
	SplayTree<int> sp;
	assert(sp.size() == 3);
	assert(sp.find(4)->value == 4);
	assert(sp.find(3)->value == 3);
	assert(sp.find(5)->value == 5);
	assert(sp.find(2) == nullptr);
	assert(sp.find(6) == nullptr);
	assert(sp.size() == 2);
	assert(sp.find(3) == nullptr);
	assert(sp.find(5)->value == 5);
	assert(sp.find(4)->value == 4);
	vector<ll> expected{-2, 4, 5, 6, 20};
	assert(sp.size() == expected.size());
	for (auto x : expected) {
		assert(sp.find(x)->value == x);
	vector<ll> vec;
	copy(sp.begin(), sp.end(), back_inserter(vec));
	assert(vec == expected);

PS: If you liked this code, you can check out my template libary OmniTemplate.

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3 years ago, # |
  Vote: I like it +9 Vote: I do not like it

Thanks for this tutorial! Do you know any good judge and/or problems to test our splay tree?

  • »
    3 years ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    You can use splay tree instead of cartesian tree. However, the main application is link cut tree.

23 months ago, # |
  Vote: I like it +8 Vote: I do not like it
template <bool dir> void advance() {
			if (node->child[1]) {
				node = extremum<!dir>(node->child[1]);
			for (; node->parent && node->side() == dir; node = node->parent)
			node = node->parent;

child[1], Is it wrong here?