At some point in your career (today?!) you will want to learn data structures. It’s not just to ace the technical interview and land your dream job. Learning data structures will help you understand how software works and improve your problemsolving skills. In this tutorial, you will learn the breadthfirst search (BFS) algorithm with graph data structures in JavaScript. If you’re just joining us, you may want to start with Learn JavaScript Graph Data Structure.
Retrieval Practice
Retrieval practice is the surest way to solidify any new learning. Attempt to answer the following questions before proceeding:

What is a Graph?

What problem(s) does a Graph solve?

What problem(s) do data structures solve?
What is a Graph?
A graph consists of a set of nodes, or vertices, connected by edges. An edge consists of a pair of vertices. For example, if we establish a pair between two vertices, A
and B
, we refer t0 this related pairing as an edge. Because they are connected by an edge, A
and B
are adjacent.
What Problem(s) Does a Graph Solve?

Optimization: We can use the graph data structure in conjunction with an optimization algorithm for determining an optimal path, such as GPS

Network topology: We can use the graph data structure when modeling network topology, such as the internet or your friends on Facebook!
What Problem(s) Do Data Structures Solve?
According to Wikipedia:
Different types of data structures are suited to different kinds of applications, and some are highly specialized to specific tasks. Data structures provide a means to manage large amounts of data efficiently for uses such as large databases and internet indexing services. Usually, efficient data structures are key to designing efficient algorithms.
Let’s Get Meta

What is BreadthFirst Search?

What is the difference between BreadthFirst Search and DepthFirst Search?

What problem(s) does BreadthFirst Search solve?
Graph Traversal: BFS vs. DFS
There are two algorithms for graph traversal:

breadthfirst search (BFS)

depthfirst search (DFS)
How do we search a graph?
Unlike a binary tree, we don’t necessarily have a root and we definitely don’t have a predetermined structure of branches. What do we have?

vertices

edges
What do we know about these things?

vertices are connected to each other with edges

vertices can be connected to any number of adjacent vertices (including zero!)
Let’s draw a simple graph so we can visualize this:
If we can’t start at a root, where do we begin?
Anywhere!
We just pick a vertex and start searching.
For the sake of simplicity, let’s choose vertex A
as our starting point, or root
, and G
as our goal
.
As we can see, A
is connected to vertices B
, C
, and D
.
Now we need to make a decision.
Do we first search the vertices connected to A
? Or do we choose one of the vertices connected to A
and then search the vertices connected to it?
🤔
This is the difference between BFS and DFS.
With BreadthFirst Search, we search all of the edges connected to a vertex before moving on to search the edges of the connected vertices.
With DepthFirst Search, we follow the paths of the edges connected to our starting vertex, or search key, one at a time, until we reach the end, then we backtrack and search the alternate paths, until we find the vertex we are looking for or we arrive back where we started.
BreadthFirst Search (BFS) in JavaScript
Let’s declare our Graph data structure.
class Graph {
constructor() {
this.vertices = [];
this.adjacent = {};
this.edges = 0;
}
addVertex(v) {
this.vertices.push(v);
this.adjacent[v] = [];
}
addEdge(v, w) {
this.adjacent[v].push(w);
this.adjacent[w].push(v);
this.edges++;
}
}
Next, let’s initialize a new Graph and add vertices and edges.
const g = new Graph();
g.addVertex("A");
g.addVertex("B");
g.addVertex("C");
g.addVertex("D");
g.addVertex("E");
g.addVertex("F");
g.addVertex("G");
g.addEdge("A","B");
g.addEdge("A","C");
g.addEdge("A","D");
g.addEdge("B","C");
g.addEdge("B","D");
g.addEdge("C","D");
g.addEdge("C","E");
g.addEdge("D","F");
g.addEdge("F","G");
To our Graph class, let’s add a bfs
method. We need to declare two parameters, goal
and root
, and if goal
and root
are equal, we return true
, else we return false
:
bfs(goal, root) {
if (root === goal) {
return true;
}
return false;
}
We can take our declaration one step further using JavaScript’s default parameters.
bfs(goal, root = this.vertices[0]) {
if (root === goal) {
return true;
}
return false;
}
📝 You will see many different implementations of BFS. Some do not specify a root while others do not specify a goal. The goal here (no pun intended) is to demonstrate an approach that covers the breadth (pun intended) of BFS variations.
Let’s verify that our bfs()
method works by logging the return value with an argument of G
:
console.log(g.bfs("G", "G"));
The above will return true
.
And the following will return false
:
console.log(g.bfs("G"));
Now what?
Let’s outline a rough approach in pseudocode:

Check the root.

If the root is equal to the goal, return true.

If the root is not equal to our goal, check the vertices adjacent to our root.

If an adjacent value is equal to our goal, return
true
. 
If none of the adjacent vertices are equal to our goal, return
false
.
If we translate this to JavaScript:
bfs(goal, root = this.vertices[0]) {
let adj = this.adjacent;
if (root === goal) {
return true;
}
for (let i = 0; i < adj[root].length; i++) {
if (adj[root][i] === goal) {
return true;
}
}
return false;
}
This will work if our query is B
, C
, or D
. How do we get to G
?
How do we “move down” a level?
Here’s the point where we need to make a jump, literally and figuratively.
But things are about to get messy! Why?
We know our graph looks likes this:
What if we don’t know what our graph looks like?
What if our graph looked like this?
Or this?
There are a couple problems we need to solve.
If we jump back to where we started, how do we avoid checking B
twice?
And…
If we checked B
, but not its adjacent vertices, how do we “move down a level” and check the vertices connected to B
?
Once we check B
, how do we move on to C
and check its adjacent vertices?
We need a way to track which vertices we already visited.
But!
We also need a way to track which vertices we need to visit.
What’s the problem we need to solve?
There’s no predetermined structure.
How do we bring order to this chaos?
🤔
Let’s restate our goal:
Given a graph, a root, and a goal, start at the root and search the adjacent vertices until we find the goal
There’s a keyword here…
It starts with ‘u’ and ends with ‘ntil’…
What is the control flow statement that functions like a repeating if
statement?
I’ll just wait here while you think about it 😛
Let’s edit our pseudocode:

While there are vertices to check, start with the root.

If the root is equal to the goal, return true.

If the root is not equal to our goal, check the vertices adjacent to our root.

If an adjacent value is equal to our goal, return
true
. 
If none of the adjacent vertices are equal to our goal, return
false
.
The next question is, while
what? How do we know there are vertices to check?
It’s like we need a list…
Let’s use an analogy.
Graphs are often used to represent social networks. Imagine you and your friends are going to see a movie. You can’t all rush into the theater at once. What do you need to do? Form a line, or, as they say across the pond, a queue!
🤯
Let’s refactor our bfs
method to use a queue. We’ll first use an array and treat it like a Queue data structure and then later refactor to use a proper Queue class.
bfs(goal, root = this.vertices[0]) {
let adj = this.adjacent;
const queue = [];
queue.push(root);
while(queue.length) {
let v = queue.shift();
if (v === goal) {
return true;
}
for (let i = 0; i < adj[v].length; i++) {
queue.push(adj[v][i]);
}
}
return false;
}
We declare an array, queue
, and push
, or enqueue our root
to it. While there are vertices in the queue, we dequeue the first vertex and assign it to the variable, v
. If v
is equal to our goal
, we return true
. If v
is not equal to our goal
, we push the adjacent vertices adjacent to our queue
and then check them.
This works great if we are only searching D
.
What happens if we search for G
?
Let’s log v
to see what’s happening inside our while
loop.
bfs(goal, root = this.vertices[0]) {
let adj = this.adjacent;
const queue = [];
queue.push(root);
while(queue.length) {
let v = queue.shift();
console.log(v);
if (v === goal) {
return true;
}
for (let i = 0; i < adj[v].length; i++) {
queue.push(adj[v][i]);
}
}
return false;
}
When we search for D
:
console.log(g.bfs("D"));
We log the following:
A
B
C
D
true
But if we search for G
:
console.log(g.bfs("G"));
We log the following:
A
B
C
D
A
C
D
A
B
D
E
A
B
C
F
B
C
D
A
B
D
E
A
B
C
F
B
C
D
A
C
D
A
B
C
F
C
B
C
D
A
C
D
A
B
D
E
D
G
true
This is not very efficient.
It gets worse. What happens if we search for H
?
Yep…
What’s the problem?
♾️
The condition of our while
loop is always true because we never stop pushing vertices to our queue.
What’s the solution?
🛑
We need an exit strategy… or a way to track which vertices we already checked so we don’t check them again.
We could use another Queue, but we don’t need FIFO, so we can just use an array to track which vertices we “discovered”.
bfs(goal, root = this.vertices[0]) {
let adj = this.adjacent;
const queue = [];
queue.push(root);
const discovered = [];
discovered[root] = true;
while(queue.length) {
let v = queue.shift();
console.log(v);
if (v === goal) {
return true;
}
for (let i = 0; i < adj[v].length; i++) {
if (!discovered[adj[v][i]]) {
discovered[adj[v][i]] = true;
queue.push(adj[v][i]);
}
}
}
return false;
}
We declare an array, discovered
, and create an index, root
assigned a value of true
. Within the for
loop in our while
loop, we add a condition to check whether or not the vertex was “discovered”. If we did not previously discover the vertex, we now mark it as discovered, or true
, and push it to our queue
to later check if it is equal to our goal
.
That’s it!
That’s BreadthFirst Search!`
You’re like, “BFD.
“Why would we need this if we can simply look up our query in the adjacent
‘dictionary’?”
I’m glad you asked.
For the sake of brevity and example, the method above is contrived. A “realworld” application of a breadthfirst search algorithm would check for a value stored in a graph and return the unique identifier, or key, of the vertex where that value was found. Another “realworld” scenario is finding the shortest path between two vertices. In the next tutorial, we’ll modify our bfs
method to do just that!
Reflection

What is BreadthFirst Search?

What is the difference between BreadthFirst Search and DepthFirst Search?

What problem(s) does BreadthFirst Search solve?
What is BreadthFirst Search?
BreadthFirst Search is an algorithm that searches a graph for a specific goal by checking all of the edges connected to a vertex before moving on to check the edges of the connected vertices.
What is the Difference Between BreadthFirst Search and DepthFirst Search?
BreadthFirst Search checks all of the vertices adjacent to a given vertex before checking the vertices adjacent to those vertices. DepthFirst Search, on the other hand, checks all of the vertices on a path and then backtracks.
What Problem(s) Does BreadthFirst Search Solve?
There are a number of specific use cases, such as the FordFulkerson or Cheney’s algorithm, for breadthfirst search algorithms, but a general application is to find the shortest, or most efficient, path between two vertices.