L = [[start_node]]+[[] for _ in 1 ... n]
start_node.visited = True
for i = 0 ... n-1:
for u in L[i]:
for v in neighbor(u):
if not v.visited:
v.visited = True
L[i+1].append(v)
This uses \(O\qty(n+m)\).
We can find the shortest paths in \(O\qty(m)\)