The edges of the Graph is given to you. g. addEdge (0, 1); g. addEdge (0, 2); g. addEdge (2, 3); g. addEdge (2, 4); g. addEdge (4, 5); g. addEdge (1, 3); g. addEdge (3, 5); Your code will need to...

Java

Very important:

As a


COMMENT IN CODE

, please DO
Test-Cases
on how you would test your solution

assumptions and hence your code

explore a specific way to perform a Breadth First Search (BFS) of a given Graph [Ref : Figure 1].

Extracted text: The edges of the Graph is given to you. g. addEdge (0, 1); g. addEdge (0, 2); g. addEdge (2, 3); g. addEdge (2, 4); g. addEdge (4, 5); g. addEdge (1, 3); g. addEdge (3, 5); Your code will need to return the traversal of the nodes in BFS order, where the traversal starts from Node/Vertex 0. When you follow the traversal process as specified - the complexity of the solu- tion will be linear as shown below. Time Complexity: 0(V + E), where V is the number of Vertices and E is the number of Edges respectively. Space Complexity: 0(V) The linear space complexity would come from the specific data structure you employ to traverse the Graph using BFS.


Extracted text: 2 4 5 Figure 1: Graph for Traversal /* Class representing a directed graph using adjacency lists */ static class Graph int V; //Number of Vertices LinkedList [] adj; // adjacency lists //Constructor Graph (int V) { this.V = V; adj - new LinkedList [V]; for (int i = 0; i < adj.length;="" i++)="" adj="" [i]="new">(); //To add an edge to graph void addEdge (int v, int w) { adj [v].add (w); // Add w to the list of v. 2