The basic algorithm for BFS:
set start vertex to visited
load it into queue
while queue not empty
for each edge incident to vertex
if its not visited
load into queue
mark vertex
So I would think the time complexity would be:
v1 + (incident edges) + v2 + (incident edges) + .... + vn + (incident edges)
where
v
is vertex 1
to n
Then the sum
v1 + (incident edges) + v2 + (incident edges) + .... + vn + (incident edges)
can be rewritten as
(v1 + v2 + ... + vn) + [(incident_edges v1) + (incident_edges v2) + ... + (incident_edges vn)]
and the first group is
O(N)
while the other is O(E).