Tree data structures have many uses, and it's good to have a basic understanding of how they work. Trees are the basis for other very used data structures like Maps and Sets. Also, they are used on databases to perform quick searches. The HTML DOM uses a tree data structure to represents the hierarchy of elements. In this post, we are going to explore the different types of trees like a binary tree, binary search trees, and how to implement them.
In the previous post, we explored the Graph data structures, which are a generalized case of trees. Let's get started learning what tree data structures are!
You can find all these implementations and more in the Github repo:
amejiarosario / dsa.jsdatastructuresalgorithmsjavascript
π₯Data Structures and Algorithms explained and implemented in JavaScript + eBook
Data Structures and Algorithms in JavaScript
This is the coding implementations of the DSA.js book and the repo for the NPM package.
In this repository, you can find the implementation of algorithms and data structures in JavaScript. This material can be used as a reference manual for developers, or you can refresh specific topics before an interview. Also, you can find ideas to solve problems more efficiently.
Table of Contents
Installation
You can clone the repo or install the code from NPM:
npm install dsa.js
and then you can import it into your programs or CLI
const { LinkedList, Queue, Stack } = require('dsa.js');
For a full list of all the exposed data structures and algorithms see.
Features
Algorithms are anβ¦
Trees: basic concepts
A tree is a data structure where a node can zero or more children. Each node contains a value. Like graphs, the connection between nodes is called edges. A tree is a type of graph, but not all of them are trees (more on that later).
These data structures are called "trees" because the data structure resembles a tree π³. It starts with a root node and branch off with its descendants, and finally, there are leaves.
Here are some properties of trees:
 The topmost node is called root.
 A node without children is called leaf node or terminal node.

Height (h) of the tree is the distance (edge count) between the farthest leaf to the root.

A
has a height of 3 
I
has a height of 0


Depth or level of a node is the distance between the root and the node in question.

H
has a depth of 2 
B
has a depth of 1

Implementing a simple tree data structure
As we saw earlier, a tree node is just a data structure that has a value and has links to their descendants.
Here's an example of a tree node:
class TreeNode {
constructor(value) {
this.value = value;
this.descendents = [];
}
}
We can create a tree with 3 descendents as follows:
// create nodes with values
const abe = new TreeNode('Abe');
const homer = new TreeNode('Homer');
const bart = new TreeNode('Bart');
const lisa = new TreeNode('Lisa');
const maggie = new TreeNode('Maggie');
// associate root with is descendents
abe.descendents.push(homer);
homer.descendents.push(bart, lisa, maggie);
That's all; we have a tree data structure!
The node abe
is the root and bart
, lisa
and maggie
are the leaf nodes of the tree. Notice that the tree's node can have a different number of descendants: 0, 1, 3, or any other value.
Tree data structures have many applications such as:
 Maps
 Sets
 Databases
 Priority Queues
 Querying an LDAP (Lightweight Directory Access Protocol)
 Representing the Document Object Model (DOM) for HTML on the Websites.
Binary Trees
Trees nodes can have zero or more children. However, when a tree has at the most two children, then it's called binary tree.
Full, Complete and Perfect binary trees
Depending on how nodes are arranged in a binary tree, it can be full, complete and perfect:
 Full binary tree: each node has exactly 0 or 2 children (but never 1).
 Complete binary tree: when all levels except the last one are full with nodes.
 Perfect binary tree: when all the levels (including the last one) are full of nodes.
Look at these examples:
These properties are not always mutually exclusive. You can have more than one:
 A perfect tree is always complete and full.
 Perfect binary trees have precisely
2^k  1\
nodes, wherek
is the last level of the tree (starting with 1).
 Perfect binary trees have precisely
 A complete tree is not always
full
. Like in our "complete" example, since it has a parent with only one child. If we remove the rightmost gray node, then we would have a complete and full tree but not perfect.
 A full tree is not always complete and perfect.
Binary Search Tree (BST)
Binary Search Trees or BST for short are a particular application of binary trees. BST has at most two nodes (like all binary trees). However, the values are in such a way that the left children value must be less than the parent, and the right children is must be higher.
Duplicates: Some BST doesn't allow duplicates while others add the same values as a right child. Other implementations might keep a count on a case of the duplicity (we are going to do this one later).
Let's implement a Binary Search Tree!
BST Implementation
BST are very similar to our previous implementation of a tree. However, there are some differences:
 Nodes can have at most, only two children: left and right.
 Nodes values has to be ordered as
left < parent < right
.
Here's the tree node. Very similar to what we did before, but we added some handy getters and setters for left and right children. Notice that is also keeping a reference to the parent and we update it every time add children.
const LEFT = 0;
const RIGHT = 1;
class TreeNode {
constructor(value) {
this.value = value;
this.descendents = [];
this.parent = null;
}
get left() {
return this.descendents[LEFT];
}
set left(node) {
this.descendents[LEFT] = node;
if (node) {
node.parent = this;
}
}
get right() {
return this.descendents[RIGHT];
}
set right(node) {
this.descendents[RIGHT] = node;
if (node) {
node.parent = this;
}
}
}
Ok, so far we can add a left and right child. Now, let's do the BST class that enforces the left < parent < right
rule.
class BinarySearchTree {
constructor() {
this.root = null;
this.size = 0;
}
add(value) { /* ... */ }
find(value) { /* ... */ }
remove(value) { /* ... */ }
getMax() { /* ... */ }
getMin() { /* ... */ }
}
Let's implementing insertion.
BST Node Insertion
To insert a node in a binary tree, we do the following:
 If a tree is empty, the first node becomes the root and you are done.
 Compare root/parent's value if it's higher go right, if it's lower go left. If it's the same, then the value already exists so you can increase the duplicate count (multiplicity).
 Repeat #2 until we found an empty slot to insert the new node.
Let's do an illustration how to insert 30, 40, 10, 15, 12, 50:
We can implement insert as follows:
add(value) {
const newNode = new TreeNode(value);
if (this.root) {
const { found, parent } = this.findNodeAndParent(value);
if (found) { // duplicated: value already exist on the tree
found.meta.multiplicity = (found.meta.multiplicity  1) + 1;
} else if (value < parent.value) {
parent.left = newNode;
} else {
parent.right = newNode;
}
} else {
this.root = newNode;
}
this.size += 1;
return newNode;
}
We are using a helper function called findNodeAndParent
. If we found that the node already exists in the tree, then we increase the multiplicity
counter. Let's see how this function is implemented:
findNodeAndParent(value) {
let node = this.root;
let parent;
while (node) {
if (node.value === value) {
break;
}
parent = node;
node = ( value >= node.value) ? node.right : node.left;
}
return { found: node, parent };
}
findNodeAndParent
goes through the tree searching for the value. It starts at the root (line 2) and then goes left or right based on the value (line 10). If the value already exists, it will return the node found
and also the parent. In case that the node doesn't exist, we still return the parent
.
BST Node Deletion
We know how to insert and search for value. Now, we are going to implement the delete operation. It's a little trickier than adding, so let's explain it with the following cases:
Deleting a leaf node (0 children)
30 30
/ \ remove(12) / \
10 40 > 10 40
\ / \ \ / \
15 35 50 15 35 50
/
12*
We just remove the reference from node's parent (15) to be null.
Deleting a node with one child.
30 30
/ \ remove(10) / \
10* 40 > 15 40
\ / \ / \
15 35 50 35 50
In this case, we go to the parent (30) and replace the child (10), with a child's child (15).
Deleting a node with two children
30 30
/ \ remove(40) / \
15 40* > 15 50
/ \ /
35 50 35
We are removing node 40, that has two children (35 and 50). We replace the parent's (30) child (40) with the child's right child (50). Then we keep the left child (35) in the same place it was before, so we have to make it the left child of 50.
Another way to do it to remove node 40, is to move the left child (35) up and then keep the right child (50) where it was.
30
/ \
15 35
\
50
Either way is ok as long as you keep the binary search tree property: left < parent < right
.
Deleting the root.
30* 50
/ \ remove(30) / \
15 50 > 15 35
/
35
Deleting the root is very similar to removing nodes with 0, 1, or 2 children that we discussed earlier. The only difference is that afterward, we need to update the reference of the root of the tree.
Here's an animation of what we discussed.
In the animation, it moves up the left child/subtree and keeps the right child/subtree in place.
Now that we have a good idea how it should work, let's implement it:
remove(value) {
const nodeToRemove = this.find(value);
if (!nodeToRemove) return false;
// Combine left and right children into one subtree without nodeToRemove
const nodeToRemoveChildren = this.combineLeftIntoRightSubtree(nodeToRemove);
if (nodeToRemove.meta.multiplicity && nodeToRemove.meta.multiplicity > 1) {
nodeToRemove.meta.multiplicity = 1; // handle duplicated
} else if (nodeToRemove === this.root) {
// Replace (root) node to delete with the combined subtree.
this.root = nodeToRemoveChildren;
this.root.parent = null; // clearing up old parent
} else {
const side = nodeToRemove.isParentLeftChild ? 'left' : 'right';
const { parent } = nodeToRemove; // get parent
// Replace node to delete with the combined subtree.
parent[side] = nodeToRemoveChildren;
}
this.size = 1;
return true;
}
Here are some highlights of the implementation:
 First, we search if the node exists. If it doesn't, we return false and we are done!
 If the node to remove exists, then combine left and right children into one subtree.
 Replace node to delete with the combined subtree.
The function that combines left into right subtree is the following:
BinarySearchTree.prototype.combineLeftIntoRightSubtree
combineLeftIntoRightSubtree(node) {
if (node.right) {
const leftmost = this.getLeftmost(node.right);
leftmost.left = node.left;
return node.right;
}
return node.left;
}
For instance, let's say that we want to combine the following tree and we are about to delete node 30
. We want to mix 30's left subtree into the right one. The result is this:
30* 40
/ \ / \
10 40 combine(30) 35 50
\ / \ > /
15 35 50 10
\
15
Now, and if we make the new subtree the root, then node 30
is no more!
Binary Tree Transversal
There are different ways of traversing a Binary Tree depending on the order that the nodes are visited: inorder, preorder and postorder. Also, we can use the DFSand BFS that we learned from the graph post. Letβs go through each one.
InOrder Traversal
Inorder traversal visit nodes on this order: left, parent, right.
BinarySearchTree.prototype.inOrderTraversal
* inOrderTraversal(node = this.root) {
if (node.left) { yield* this.inOrderTraversal(node.left); }
yield node;
if (node.right) { yield* this.inOrderTraversal(node.right); }
}
Let's use this tree to make the example:
10
/ \
5 30
/ / \
4 15 40
/
3
Inorder traversal would print out the following values: 3, 4, 5, 10, 15, 30, 40
. If the tree is a BST, then the nodes will be sorted in ascendent order as in our example.
PostOrder Traversal
Postorder traversal visit nodes on this order: left, right, parent.
BinarySearchTree.prototype.postOrderTraversal
* postOrderTraversal(node = this.root) {
if (node.left) { yield* this.postOrderTraversal(node.left); }
if (node.right) { yield* this.postOrderTraversal(node.right); }
yield node;
}
Postorder traversal would print out the following values: 3, 4, 5, 15, 40, 30, 10
.
PreOrder Traversal and DFS
Inorder traversal visit nodes on this order: parent, left, right.
BinarySearchTree.prototype.preOrderTraversal
* preOrderTraversal(node = this.root) {
yield node;
if (node.left) { yield* this.preOrderTraversal(node.left); }
if (node.right) { yield* this.preOrderTraversal(node.right); }
}
Preorder traversal would print out the following values: 10, 5, 4, 3, 30, 15, 40
. This order of numbers is the same result that we would get if we run the DepthFirst Search (DFS).
BinarySearchTree.prototype.dfs
* dfs() {
const stack = new Stack();
stack.add(this.root);
while (!stack.isEmpty()) {
const node = stack.remove();
yield node;
// reverse array, so left gets removed before right
node.descendents.reverse().forEach(child => stack.add(child));
}
}
If you need a refresher on DFS, we covered in details on Graph post.
BreadthFirst Search (BFS)
Similar to DFS, we can implement a BFS by switching the Stack
by a Queue
:
BinarySearchTree.prototype.bfs
* bfs() {
const queue = new Queue();
queue.add(this.root);
while (!queue.isEmpty()) {
const node = queue.remove();
yield node;
node.descendents.forEach(child => queue.add(child));
}
}
The BFS order is: 10, 5, 30, 4, 15, 40, 3
Balanced vs. Nonbalanced Trees
So far, we have discussed how to add
, remove
and find
elements. However, we haven't talked about runtimes. Let's think about the worstcase scenarios.
Let's say that we want to add numbers in ascending order.
We will end up with all the nodes on the left side! This unbalanced tree is no better than a LinkedList, so finding an element would take O(n). π±
Looking for something in an unbalanced tree is like looking for a word in the dictionary page by page. When the tree is balanced, you can open the dictionary in the middle and from there you know if you have to go left or right depending on the alphabet and the word you are looking for.
We need to find a way to balance the tree!
If the tree was balanced, then we could find elements in O(log n) instead of going through each node. Let's talk about what balanced tree means.
If we are searching for 7
in the nonbalanced tree, we have to go from 1 to 7. However, in the balanced tree, we visit: 4
, 6
, and 7
. It gets even worse with larger trees. If you have one million nodes, searching for a nonexisting element might require to visit all million while on a balanced tree it just requires 20 visits! That's a huge difference!
We are going to solve this issue in the next post using selfbalanced trees (AVL trees).
Summary
We have covered much ground for trees. Let's sum it up with bullets:
 The tree is a data structure where a node has 0 or more descendants/children.
 Tree nodes don't have cycles (acyclic). If it has cycles, it is a Graph data structure instead.
 Trees with two children or less are called: Binary Tree
 When a Binary Tree is sorted in a way that the left value is less than the parent and the right children is higher, then and only then we have a Binary Search Tree.
 You can visit a tree in a pre/post/inorder fashion.
 An unbalanced has a time complexity of O(n). π€¦π»β
 A balanced has a time complexity of O(log n). π
Discussion (2)
Trees are fun π²π²π². I like your diagrams and am looking forward to the AVL post.
I had a couple of questions / notes:
I'm confused about where you said that the DOM is a binary tree, since DOM nodes can have any number of children. (Specifically you said "searching the DOM" but it seems like the same story?)
Also, in the same section, your "map" bullet links to your article saying:
I get your meaning but tend to think of it more like:
Maps, dictionaries, and associative arrays all describe the same abstract data type. But hash map implementations are distinct from tree map implementations in that one uses a hash table and one uses a search tree. (With the differences in sort order and time complexity that those two imply.)
In other words, when I read "hash map," I think "implemented using a hash table." My attitude may be overly influenced by the Java standard library, but it seems like a fair assumption, and I can't find any counterexamples at the moment.
Thanks, John, for your comments.
Agree that the DOM is not a binary tree, but a tree. I doubled check the intro, and it looks correct. However, I listed the tree applications in the wrong place. It should be outside the binary tree section. Thanks for pointing that out, I'll fix it.
About the Hashmap article, that's right. I could have been more descriptive. I'll make the change. BTW, I implemented HashMaps and also TreeMaps in JavaScript with a similar interface used in the Java standard API. It's not covered on the posts but is on the DSA.js book and github repo: github.com/amejiarosario/dsa.js/bl...
About the AVL tree post you can find it in my blog already or wait for next week when I publish it here ;)