algorithm - Is the worst time complexity of BFS in a graph traversal n+2E? -
i understand time complexity of bfs in graph traversal o( v + e ) since every vertex , every edge explored in worst case.
well,is exact time complexity v+2e ??
every vertex explored once+ every adjacent vertices
the sum of degree of vertices in graph= no of edges*2= 2e
thus time complexity n+2e..am correct?
for random graph, time complexity o(v+e): breadth-first search
as stated in link, according topology of graph, o(e) may vary o(v) (if graph acyclic) o(v^2) (if vertices connected each other).
therefore time complexity varies fromo(v + v) = o(v) o(v + v^2) = o(v^2) according topology of graph.
besides, since |v| <= 2 |e|, o(3e) = o(e) correct, bound looser.
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