bbst

v0.2.1

Published

2 years ago

Balanced binary search tree implemented as Treaps

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johnsonzhong

binary search tree fast fast iteration iteratable self-balancing Treap

Balanced binary search trees in ES6 Javascript for Node.js and browsers

The underlying implementation is a Treap, which provides balance with high probability. This implementation focuses on fast iteration, and provides generators for doing so.

Treaps are simple and fast. In most cases, it provides up to 4x faster inserts, and 2x faster erases than a similar RB tree implementation.

Deterministic behaviour can be achieved by seeding the RNG.

Installation

Package name is bbst

npm install bbst --save

Test

npm install mocha

// perform random test
npm test

// test with specific seed "seed"
npm test seed

Usage

The API provides insert, changeKey, find, erase, iteration, reverseIterator, size, height, print, and variants of the above (to be covered in examples below).

First you need to include the module, which is easy in both node and a browser (look at index.html).

node

const BST = require("bbst");

browser

<html>
<head>
    <meta charset="UTF-8">
    <script src="src/treap.js"></script>
</head>
<body>
<script>
const BST = Treap;
</script>
</body>
</html>

From this point on, usage is the same on both platforms; let's assume we name the class BST.

deterministic seed

BST.seedrandom("some value");

construction

const bst = new BST();

insert

Insert takes 1 object that is required to have a key property that can be compared. An Error will be thrown if something is inserted without a key. The inserted object can have any other property.

bst.insert({key:1, name:"Buzz Lightyear"});

Duplicate keys are not guarenteed to remain in insertion order.

bst.insert({key:1, name:"An imposter"});
// bst.find(1) will find them in a random order

iteration

Iterating forward will do so in increasing key order

for (let node of bst) {
  console.log(node.key);  // print keys smallest to largest
}
for (let node of bst.reverseIterator()) {
  console.log(node.key);  // print keys largest to smallest
}

This is very useful for game tree searches as you iterate through possible moves in place rather than creating and sorting them at each step. This is especially powerful combined with changeKey. For each step, instead of creating a new state, simply change the current one to the next one while tracking the changes. If you have n possible moves, at each step you use O(1) memory instead of O(n).

I use this in a minimax algorithm with alpha-beta pruning so that the best moves are explored first, which often causes most of the other moves to be pruned and thus not generated.

changeKey

A small change to a key's value can be dealt with faster than deleting and reinserting the node with changeKey.

const bst = new BST();
for (let key = 0; key < 1000000; ++key) {
  bst.insert({key});
}
const node = bst.find(4000);
// takes many fewer operations since it searches from node up
bst.changeKey(node, 4001);

print

The tree comes with pretty printing that works in consoles and browsers, as well as returning the string.

const bst = new BST();
for (let key = 0; key < 10; ++key) {
    bst.insert({key});
}
bst.print();

Results in something similar to

4
├ 1
│ ├ 0
│ └ 2
│   ├ 
│   └ 3
└ 9
  └ 7
    ├ 5
    │ ├ 
    │ └ 6
    └ 8

print accepts a custom printer function for printing more than just the node key.

bst.print((node)=> {
  // return something printable, such as a string
  return node.key + ":" + node.name;
});

find

All failed finds will return BST.NIL. The basic find retrieves the top-most node with matching key.

const node = bst.find(5);
const nonNode = bst.find(11);
assert(nonNode === BST.NIL);  // true

If there is temporal locality (you need it soon after you found it, but not immediately after), then findAndElevate will make subsequent finds faster.

const hotNode = bst.find(5);

If your data has duplicate keys and you care about finding all of them or a specific one, then use findFirst and findNext. findFirst takes the key to find, findNext takes the previous node (and is O(1) time).

bst.insert({key:5, fruit:"Pomelo"});
bst.insert({key:5, fruit:"Tomato"});
let found = bst.findFirst(5);
while (found !== BST.NIL) {
  console.log(found.fruit);
  found = bst.findNext(found);
}

erase

Erase the top-most node with a matching key, or do nothing if it doesn't exist

bst.erase(5);

Erase a specific node, which is useful for duplicates and faster if you already found the node

bst.erase(bst.find(5));

size

Get number of nodes

let bst = new BST();
bst.size(); // 0

for (let key = 0; key < 10; ++key) {
  bst.insert({key});
}
bst.size(); // 10

height

Maximum pointer distance to root node. It is expected to be O(lg(bst.size())), and is more likely with larger size.

let bst = new BST();
bst.height(); // 0


for (let key = 0; key < 1000; ++key) {
  bst.insert({key});
}
bst.height(); // about lg(bst.size())