- Trending Categories
- Data Structure
- Networking
- RDBMS
- Operating System
- Java
- iOS
- HTML
- CSS
- Android
- Python
- C Programming
- C++
- C#
- MongoDB
- MySQL
- Javascript
- PHP

- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who

Graph traversal is the problem of visiting all the vertices of a graph in some systematic order. There are mainly two ways to traverse a graph.

- Breadth First Search
- Depth First Search

**Breadth First Search (BFS)** starts at starting level-0 vertex X of the graph G. Then we visit all the vertices that are the neighbors of X. After visiting, we mark the vertices as "visited," and place them into level-1. Then we start from the level-1 vertices and apply the same method on every level-1 vertex and so on. The BFS traversal terminates when every vertex of the graph has been visited.

**BFS Algorithm**

The concept is to visit all the neighbor vertices before visiting other neighbor vertices of neighbor vertices.

Initialize status of all nodes as “Ready”.

Put source vertex in a queue and change its status to “Waiting”.

Repeat the following two steps until queue is empty −

Remove the first vertex from the queue and mark it as “Visited”.

Add to the rear of queue all neighbors of the removed vertex whose status is “Ready”. Mark their status as “Waiting”.

**Problem**

Let us take a graph (Source vertex is ‘a’) and apply the BFS algorithm to find out the traversal order.

**Solution** −

Initialize status of all vertices to “Ready”.

Put

*a*in queue and change its status to “Waiting”.Remove

*a*from queue, mark it as “Visited”.Add

*a*’s neighbors in “Ready” state*b, d*and*e*to end of queue and mark them as “Waiting”.Remove

*b*from queue, mark it as “Visited”, put its “Ready” neighbor*c*at end of queue and mark*c*as “Waiting”.Remove

*d*from queue and mark it as “Visited”. It has no neighbor in “Ready” state.Remove

*e*from queue and mark it as “Visited”. It has no neighbor in “Ready” state.Remove

*c*from queue and mark it as “Visited”. It has no neighbor in “Ready” state.Queue is empty so stop.

So the traversal order is −

a → b → d → e → c

The alternate orders of traversal are −

a → b → e → d → c

Or, a → d → b → e → c

Or, a → e → b → d → c

Or, a → b → e → d → c

Or, a → d → e → b → c

**Application of BFS**

- Finding the shortest path
- Minimum spanning tree for un-weighted graph
- GPS navigation system
- Detecting cycles in an undirected graph
- Finding all nodes within one connected component

**Complexity Analysis**

Let G(V, E) be a graph with |V| number of vertices and |E| number of edges. If breadth first search algorithm visits every vertex in the graph and checks every edge, then its time complexity would be −

O( | V | + | E | ). O( | E | )

It may vary between O(1) and O(|V2|)

- Related Questions & Answers
- Breadth-first search traversal in Javascript
- Breadth First Search (BFS) for a Graph
- Breadth First Search on Matrix in C++
- Breadth First Search on Matrix in Python
- Breadth First Search or BFS for a Graph
- Web crawling using Breadth First Search at a specified depth
- Best First Search (Informed Search)
- Depth First Search
- Depth-first search traversal in Javascript
- Depth First Search (DFS) for a Graph
- Depth First Search or DFS for a Graph
- Depth-First Search on a Digraph in Data Structure
- Python Program for Depth First Binary Tree Search using Recursion
- Search for documents matching first item in an array with MongoDB?
- Python Program to Implement Depth First Search Traversal using Post Order

Advertisements