Sorted Maps in JavaScript

This time we have a guest post by Monmohan Singh. He is the author of dsjslib, a collection of data structures.

JavaScript objects themselves can serve as Maps. There are some caveats but by and large it is possible to use plain Objects as Maps in JavaScript.

However a JavaScript object by default can’t be used as a ordered Map (ordered on keys) . ECMAScript specification doesn’t define any explicit ordering of keys (while enumerating or otherwise). Most Browsers iterate over properties in the order they were created.

Sorted Maps are a data structure available in other languages (example TreeMap class in java collections) Note that the ordering happens on dynamic data set. This means that the ordering is maintained irrespective of new insertions or deletions to the Map. Compare that to say using JavaScript Array sort() which is static i.e additions/deletions of elements in the array will not maintain order or require repeated sorts to maintain order (which would roughly be O(N*logN) cost everytime) This structure can be used in applications where the need is to

In this article, we will discuss in detail about AVL Tree implementation in JavaScript which can be used create a sorted Map on dynamic data set. It is taken from dsjslib, a collection of some standard data structures (like sorted map) not available in JavaScript.

AVLTree in dsjslib is a Sorted Map with O(log N) access time for insert, delete and find operations . Also provides listing the sorted keys and values in O(N) time.

The article assumes some familiarity with Binary Search Tree data structure

Quick Look at Binary Search Trees (BST)

A binary search tree is a binary (each node has maximum of two children) tree with the additional property that for each node, the key is greater than the key of its left child and less than the key of its right child. A simple but powerful data structure. Below is a binary search tree built from the following sequence of keys: 15, 11, 21, 17, 13, 9

Basic Tree

BST allow insert, delete and search of a key in order of height of the tree (O(h)). Searching basically involves comparing key in each node and then deciding to go left (if key is less than the node or right if key is greater) .. This means that the number of comparisons would less than or equal to the longest path from root to leaf (or in other words the height of the tree). So as long as the height of the tree is kept small we are good. In simple BSTs height depends on the sequence of keys being inserted.

Here is a BST for following sequence of keys 9, 11, 13, 15, 17, 21

Basic Tree

It’s not a tree anymore but a linked list!! Now insert, delete and search take order of number of elements in the tree.

AVLTree - A balanced search Tree

Unlike a simple binary search tree, an AVL tree introduces additional constraints which force it to be a balanced tree after every operation - so, there are no worst case scenarios as you get in a binary tree.

We will discuss the AVLTree implementation in dsjslib.

So lets take the same example and see what we can do to avoid a tree becoming a linked list. Lets start building a BST and add 9, 11, 13

Basic Tree

At this stage, what if we pull 9 node down and make 11 the root, the tree looks like

Basic Tree

Its still a binary search tree and also one in which the height of the tree is less than the earlier . This is the idea behind AVL Trees to rotate the tree at certain nodes on insertion and deletion and maintain low height in the tree which guarantees O(logN) performance.

Maintaining the balance

AVLTrees maintain balance in similar way i.e. rotating the tree

Formal Notions

  1. Height of a node is defined as following
    • Non-existent nodes have height -1
    • Leaf nodes have height 0
    • A non-leaf node height = Max{leftNodeHeight,rightNodeHeight}+1 Example depicted below

Basic Tree

  1. The AVL Tree node invariant says that for any node, the difference between the left and right subtree height can’t be more than 1. This leads to maintaining the height O(log Number of nodes)

Steps are

  1. After insertion or deletion find the node where violation occurs
  2. Once we have the node which was inserted (or parent of the node deleted), move up the tree checking the height difference until we a find node where violation occurs (height difference between subtrees is more than 1).
  3. In dsjslib.AVLTree, this is taken care by the function checkAVLProperty(node). This function is invoked post insertion or deletion of a node

Consider the implementation below:

// do simple BST insert
AVLTree.prototype.put = function(key, value) {
    var ins =, key, value);

    try {
        //find where is the violation
    } catch (vErr) {
        //rebalance the tree at that node

    return ins;

Checking AVLProperty involves checking the difference in height between left and right subtree. The height is recursively (pre) calculated. Implementation below:

AVLTree.prototype.checkAVLProperty = function (node) {
    if (!node) return;

    var lc = node.leftChild, rc = node.rightChild;
    var hdiff = (rc ? rc.height : -1) - (lc ? lc.height : -1);

    if (Math.abs(hdiff) > 1) {
        if (debug)console.log("AVL Height violation at Node key" + node.key);

        throw {'node':node, 'hdiff':hdiff};


Fixing the Balance

Once we have found the node where violation occurs we need to fix that. There are 4 cases to consider

  1. Right-Right heavy

AVL Tree

The node where the violation occurs (9, shown in Red) is right heavy. Its right child height is more than its left child height) AND the child (11) is also right heavy.
This can be fixed by single LEFT rotation at violated node (9)

AVL Tree

  1. Left-Left heavy

Symmetrical to right-right heavy case and right rotation at violated node fixes it

  1. Right-Left heavy

This case is little bit trickier. In this case (example shown below), the violated node is right heavy but the right child is left heavy (its left child height is -1, right child height is 0 which is greater than -1)

AVL Tree

Lets see if we can fix this by applying our earlier knowledge and rotating left at violated node .


AVL Tree

We can see that it doesn’t fix the problem. Now the violation occurs at the new parent i.e 11

These cases are fixed by two rotations

i. First rotate right at the left heavy child(11)

AVL Tree

ii. Now this is a know case of right-right heavy, lets rotate left at the violated node to balance it

AVL Tree

  1. Left-Right heavy

Is symmetric to case c) above and in this case the violated node is left heavy but its left child is right heavy. Applying the same logic as above, it will be fixed by, first a left rotation of the left child and then a right rotation at the violated node.

Implementation of Rotation
AVLTree.prototype.rebalance = function(vError) {
    var balance = vError.hdiff, vNode = vError.node;
    var child = balance > 1 /right heavy/ ? vNode.rightChild : vNode.leftChild;

    //+ve, right heavy, -ve left heavy
    var childBalance = this._nodeHeight(child);

      node is right heavy but child is left heavy and vice-versa
      @type {Boolean}
    var zigzag = balance > 1 ? childBalance < 0 : childBalance > 0;

    / Requires double rotation /
    if (zigzag) {
        //rotate on child first
        this.rotate(child, childBalance > 0? 'l': 'r')

    //rotation on node where violation occurs
    this.rotate(vNode, balance > 1? 'l': 'r');