Why is the time complexity of both DFS and BFS O( V + E )

Your 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).

Very simplified without much formality: every edge is considered exactly twice, and every node is processed exactly once, so the complexity has to be a constant multiple of the number of edges as well as the number of vertices.

DFS(analysis):

• Setting/getting a vertex/edge label takes O(1) time
• Each vertex is labeled twice
  • once as UNEXPLORED
  • once as VISITED
• Each edge is labeled twice
  • once as UNEXPLORED
  • once as DISCOVERY or BACK
• Method incidentEdges is called once for each vertex
• DFS runs in O(n + m) time provided the graph is represented by the adjacency list structure
• Recall that Σv deg(v) = 2m

BFS(analysis):

• Setting/getting a vertex/edge label takes O(1) time
• Each vertex is labeled twice
  • once as UNEXPLORED
  • once as VISITED
• Each edge is labeled twice
  • once as UNEXPLORED
  • once as DISCOVERY or CROSS
• Each vertex is inserted once into a sequence Li
• Method incidentEdges is called once for each vertex
• BFS runs in O(n + m) time provided the graph is represented by the adjacency list structure
• Recall that Σv deg(v) = 2m