Breadth First Search time complexity analysis

Question

The time complexity to go over each adjacent edge of a vertex is, say, O(N), where N is number of adjacent edges. So, for V numbers of vertices the time complexity becomes O(V*N) = O(E), where E is the total number of edges in the graph. Since removing and adding a vertex from/to a queue is O(1), why is it added to the overall time complexity of BFS as O(V+E)?

Answer 1

I hope this is helpful to anybody having trouble understanding computational time complexity for Breadth First Search a.k.a BFS.

Queue graphTraversal.add(firstVertex);  // This while loop will run V times, where V is total number of vertices in graph. while(graphTraversal.isEmpty == false)      currentVertex = graphTraversal.getVertex();      // This while loop will run Eaj times, where Eaj is number of adjacent edges to current vertex.     while(currentVertex.hasAdjacentVertices)         graphTraversal.add(adjacentVertex);      graphTraversal.remove(currentVertex); 

Time complexity is as follows:

V * (O(1) + O(Eaj) + O(1)) V + V * Eaj + V 2V + E(total number of edges in graph) V + E 

I have tried to simplify the code and complexity computation but still if you have any questions let me know.