Cartesian tree or treap (binary search tree + binary heap) is a fast yet simple data structure. It conforms to a core search binary tree property and binary heap property at the same time. Despite its simplicity, treap self-balances, resulting in O(logn)
complexity on average for all common operations.
Amazing, right?
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The algorithm uses random values. Therefore, O(logn)
complexity is on average. However, with a lot of items O(logn)
is almost always true. So, later in this article, I will use just O(logn)
without "on average" addition.
Moreover, there is a modification (implicit treap, treap with implicit key) that lets you use treap as a usual array with O(logn)
random insertions and random deletions. Isn't it cool? In this article I'll explain how to create one and provide the implementation in Swift. Also, I will compare treap to the general set
from standard library. Let's start!
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In a binary search tree, for each node, all items' values in the left subtree are less than the node's value, and all items in the right subtree are greater
Core algorithm
As I said earlier, treap combines heaps and binary search trees. Therefore, we are going to store at least two properties: key
(or value) and priority
. Key is a value for which tree is a search tree and for the priority, it is a binary heap.
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A binary heap is a binary tree where each node child's value is less than the node's value
![Treap: The Easiest Search Tree (Explained) | Alex Dremov (1) Treap: The Easiest Search Tree (Explained) | Alex Dremov (1)](https://i0.wp.com/alexdremov.me/content/images/2022/04/treap.png)
On the image above, you may notice that for every node, all child's priorities are less. On the other side, all children on the left have a key less than that in the node, and all children on the right have a larger key.
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It's also called a cartesian tree as it can be displayed on a regular 2D grid with (key, priority) coordinate for each node. Just like in the image above.
To create a fully-functioning search tree, we need to implement:
- find
- insert
- remove
More exotic operations like lower bound
and upper bound
are also pretty simple and does not differ from those in the other search trees. And all these operations can be implemented using just two helper operations!
How to do that?
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Split
Splits the tree into two trees by given value
. All values in the left tree are less than the value
while in the right tree are greater. And both resulting trees are correct treaps.
We will use a special flag that decides whether to send values that are equal to the left tree or to the right tree.
![Treap: The Easiest Search Tree (Explained) | Alex Dremov (2) Treap: The Easiest Search Tree (Explained) | Alex Dremov (2)](https://i0.wp.com/alexdremov.me/content/images/2022/04/split-image-1.png)
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Merge
Merges two treaps into one big treap.
Prerequisite: all items in the first tree are less than items in the right tree.
![Treap: The Easiest Search Tree (Explained) | Alex Dremov (3) Treap: The Easiest Search Tree (Explained) | Alex Dremov (3)](https://i0.wp.com/alexdremov.me/content/images/2022/04/merge-example.png)
So, if we implement these two methods, implementing all other three operations would be trivial.
Split
Let's start thinking about code at this stage. I'm going to explain this in C++
. Rewriting the following code in Swift
is actually really easy. Leave a comment bellow if you need a help.
template<typename T>struct Node {T key;size_t prior;Node* left = nullptr, *right = nullptr;Node(T key, size_t prior) :key(std::move(key)),prior(prior) {}};
For split, we have a head
node and a key
for which split needs to be done. This method is extremely simple using recursion.
Algorithm
Let the current head be p
.
- If
p->key
is less than thekey
, then we need to go right and splitp->right
further.Also, splitting
right
will bring two trees as well, and the first one will have nodes with keys less than thekey
. Yet, they are greater than thep->key
(as they are in the second tree of the first split).
So, we setp->right
to the first tree of splittingright
result.Result:
p
, split right's second tree - If the
p->key
is greater or equal to thekey
, then we need to go left and splitp->left
further.Similarly to the case above, we set
p->left
to the second tree of split left.Result: split left's first tree,
p
The algorithm above leaves a node that is equal to the split value in the second tree. Symmetrically, we will use the equalOnTheLeft
flag to leave the node in the left tree.
So, the final code:
pair<Node *, Node *> split (Node *p, const T& key,bool equalOnTheLeft=false) { if (!p) // reached leaf return {nullptr, nullptr}; if (p->key < key || (equalOnTheLeft && p->key == key)) { // splitting right auto q = split(p->right, key, equalOnTheLeft); // q.first has nodes of the right // subtree that are less than key p->right = q.first; return {p, q.second}; } else { // splitting left auto q = split(p->left, key, equalOnTheLeft); // q.second has nodes of the left // subtree that are greater or equal // to the key p->left = q.second; return {q.first, p};}}
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Priorities are not used and not changed during the split procedure. The resulting trees have the right order of priorities as the initial tree had it right
Merge
Merge is similar to split, but it uses priorities to do the work. As I mentioned before, there is a prerequisite: all items in the first merged tree must be less than items in the second tree. If this is not true, another algorithm must be used.
Algorithm
Similarly to split
, merge
is also recursive. Let us have two trees to merge: l
and r
.
- We need to choose which tree will represent the new head. That's simple — the head must have the greatest priority, so we choose
l
orr
based on that.
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Notice that the head node in l
has the highest priority in the whole l
tree as its a property of correct treap. The same applies to r
.
- If
l
has greater priority, thenl->left
subtree will remain intact as left subtree for sure less thanr
and it has nothing to do with it.Then,
l->right
subtree must be merged withr
and it's going to be the newl->right
subtree. - If
r
has greater priority, then, similar to the example above,r->right
will remain intact andr->left
must be merged withl
Node* merge (Node *l, Node *r) { if (!l) // left is empty return r; if (!r) // right is empty return l; if (l->prior > r->prior) { // l has the new head. l->right = merge(l->right, r); return l; } else { // r has the new head. r->left = merge(l, r->left); return r; }}
Why is it correct?
It seems like nothing stops us from breaking the search tree structure where all items' values in the left subtree are less than the node's value, and all items in the right subtree are greater.
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Prerequisite saves binary search tree property as items are never reordered and l < r
property is always kept the same
Implementing search tree methods
You believed me that all methods are easy to implement through split
and merge
. Time to prove that.
Find
Find is implemented just like for the general search tree. We use the fact that keys in the left subtree are greater than the value in the node.
Node* find(Node* node, const T& key) {if (node == nullptr)return nullptr; if (node->key == key)return node; return find(key >= node->key ? node->right : node->left, key);}
Insert
Let's think about insert in terms of split and merge. We have one big tree and we need to insert a new key
.
- Split the tree by
key
to new trees:first
andsecond
. Then, we will have two trees: the first (which has values lower than thekey
) and the second (which has values greater or equal to thekey
).We can check that node already exists: try to find it in the right tree.
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Implementation requires that each item is met only once.
If you need to insert multiple copies of the same item, you can store an item and it's count to achieve that
- Create a new node that will store the new
key
—newNode
. Ta-da this node is a correct treap that has only one node.For the new node, you need to set a random priority
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Random priorities are key to the complexity. This makes the cartesian tree balance itself, making O(logn)
complexity for all operations
- New head will be
merge(first, merge(newNode, second))
See? It's that simple.
![Treap: The Easiest Search Tree (Explained) | Alex Dremov (4) Treap: The Easiest Search Tree (Explained) | Alex Dremov (4)](https://i0.wp.com/alexdremov.me/content/images/2022/04/merge-example-specific.png)
Node* insert(Node* head, T key) { auto split = split(head, key); if (find(split.second, key) != nullptr) { // Key exists already // Merge back return merge(split.first, split.second); } auto newNode = new Node(std::move(key), rand()); return merge(split.first, merge(newNode, splitsplitted.second));}
Remove
It's very similar to insert
. However, that's where the equalOnTheLeft
flag is used.
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Remember that the second
tree produced by split
contains items greater or equal to the selected key
Therefore, the second
tree will contain the value that needs to be removed. But how to remove it from the tree?
Split again.
We can split the second
tree by key, setting the equalOnTheLeft
flag to true
. Thus, the node will be separated from the second
tree to the new tree.
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After conducting two splits and separating deleted node, unneded node is easely removed everything else is merged.
![Treap: The Easiest Search Tree (Explained) | Alex Dremov (5) Treap: The Easiest Search Tree (Explained) | Alex Dremov (5)](https://i0.wp.com/alexdremov.me/content/images/2022/04/remove-example.png)
Node *remove(Node *head, const T &key) { auto split = split(head, key); if (split.second) { auto secondSplit = split(split.second, key, /*equalOnTheLeft=*/true); // Key exists, so delete it and merge auto everythingElse = secondSplit.second; if (secondSplit.first == nullptr) { // There's no element equal to key. Merge back. return merge(split.first, everythingElse); } // We got node with key value in // secondSplit.first delete secondSplit.first; size--; return merge(split.first, everythingElse); } // Key is not presented. Merge back. return merge(split.first, split.second);}
Full code
You can download C++ code of a little bit optimized Treap here:
Comparing to std::set
First of all, the implemented version of treap utilizes split
and merge
methods. Note that there is more efficient implementation that uses rotations. However, the true power of treap is in split
and merge
methods as other search trees can't do it easily.
Find tests
It's visible that asymptotics is similar. Though, treap always has the greater overhead. Still, it's a good result! We're competing with an utterly optimized standard library data structure.
Inserts
Insertion has even bigger overhead. And it was expected: recursive calls of merge and split do not improve performance ;)
Comparison conclusion
As you see, treap has higher nodes' height on average than that of very well-balanced AVL tree.
Yes, treap has worse performance than that of std::set
. Yet, the results are comparable, and with a large data size, treap gets closer and closer to std::set
which in fact is a red and black tree.
Believe me, you don't want to write your own RB tree. It's a nightmare.
Use cases and modifications
We developed this data structure not just to lose std::set
. There are several useful applications.
Sum of numbers in the interval
We need to modify Node
structure, adding sum
field. It will store sum of all its children and itself.
template<typename T>struct Node {T key;size_t prior;long long sum;Node* left = nullptr, *right = nullptr;Node(T key, size_t prior) :key(std::move(key)),prior(prior) {}};
It's extremely easy to update the sum
. Every time childs are changed, sum = left->sum + right->sum
. So, you can implement some kind of update
function and call it in split and merge right before returning value. That's it.
How to answer on request?
We receive interval [l, r]
. To calculate the sum of numbers on this interval, we can split the tree by l
, then split the second tree of the result by r+1
(or by r
, leaving equal elements on the left). In the end, we will have a tree containing all added numbers in the interval [l, r]
.
Complexity: O(logn)
versus O(n)
naive.
Using a hash of value in place of priority
You can use a hash of value as a priority as a good hash function is pretty random. What benefits does it bring?
If keys and priorities are fixed, then no matter how you construct the treap or add elements, it's always going to have the same structure.
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You may think about it this way: keys fix x axis and priorities fix y axis of treap
Therefore, you can compare two sets in O(n)
as treaps containing the same values will have absolutely the same structure.
Implicit treap
What if we use the size of the left subtree as a key? Then, we can use this key as an index. Wow. That means that we can represent a regular ordered array as a treap!
By doing this, we can:
- make insertions by random index
O(logn)
versusO(n)
naive - make deletions by random index
O(logn)
versusO(n)
naive
With great power, comes great responsibility.
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Access by random index downgrades to O(logn)
versus O(1)
in the standard array.
If your algorithm requires a lot of array modifications and very few accesses/outputs, then it's the right choice. Moreover, you can convert treap into an array and back with O(n)
complexity.
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I have implicit treap implemented in Swift. It behaves just like the general array and implements a lot of optimisations. Check it out!
Cut-paste problem
Imagine that you have a big string and you recieve requests to cut some part and to insert it somwhere.
This problem can be solved using treaps with implicit key. You can use splits to cut needed part and merge to insert it.
FAQ
Cartesian trees are most suitable for what?
Treap is useful when you need to collect some kind of characteristic on an interval (for example, sum) or apply some modification to the interval. Treap with implicit key is also useful when you need to apply a lot of random tree insertions/deletions with few accesses.
Why don't we use array indices as keys for an implicit treap?
Because in case of insertion we would need to recalculate all indeces that are higher than inserted index. Therefore, it downgrades complexity to O(n)
.
Is treap a randomized tree?
Yes, it is. But it can also use hash value in place of a random value.
I know about implementation without split and merge. It utilizes left and right turns. Is it better?
For example, GeeksforGeeks use such implementation, I know. But I believe that the true value of treap is in seampless splits and merges. You've already seen by examples how it is really usefull. Why implementing treap with turns when you can build AVL that's probably going to be faster?
Love data structures?
Check out my article on the amazing Skip List! While a lot of people never heard about it, Skip List is beautiful and can solve, for example, the problem of finding the n-th maximum or the rolling median problem in the most efficient way.
Also, you can check the whole algorithms section of my blog
References
Introduction To Algorithms
The first edition won the award for Best 1990 Professional and Scholarly Book in Computer Science and Data Processing by the Association of American Publishers.There are books on algorithms that are rigorous but incomplete and others that cover masses of material but lack rigor. Introduction to Algo…
Google Books
https://www.cs.cmu.edu/~scandal/papers/treaps-spaa98.pdf