Bellman Ford

Detect negative cycles by exploiting the fact that after $N-1$ iteration, all paths have been traversed and we have reached found out the optimum path value from src to dst. If any value continues to updated after $N-1 $ iteration, we know that there is a negative cycle because a path is at most N-1 edges long, if a path's value is updated within this iteration, we must have traversed some cycle with negative edges.
Bellman-ford is $O(V*E)$

Level by level traversal (BFS)
The goal here is looking over all edges over an $N-1$ iteration scheme, to update vertices only if in the previous iteration it has been reached (dist != inf).
Each iteration, we simply look at the edges from u to v and update if the w of that edge affects our path costs so far.
An $N-1$ iteration scheme guarantees traversing through all paths, hence minimizing or maximizing the cost.
An extra iteration, Nth is added to detect cycles for negative edges, this is because TODO

Depending on the question, we might need to change the template algorithm a little bit.

dist[src] = 0 
for u,v,w in edges : 
    if dist[u] != inf && dist[u] + w < dist[v]: 
        dist[v] = dist[u]+w
return dist[dest] 