I don't understand for potential cyclic graph traversals, why for is there a connection between source and destination node in the graph:

i) DFS, if a node is visited, we return false

ii) BFS, if a node is visited, we continue (in the loop)

example code (from https://structy.net/problems/undirected-path):

const undirectedPath = (edges, nodeA, nodeB) => {
  const graph = buildGraph(edges);
  return hasPath(graph, nodeA, nodeB, new Set());
}

// BFS
const hasPath = (graph, src, dst, visited) => {
  const queue = [src];
  
  while(queue.length > 0){
    const current = queue.shift();
    if(current === dst) return true;
    // if it's DFS, do not "continue", instead "return false" - why?
    if(visited.has(current)) continue;
    visited.add(current);
    for(let neighbor of graph[current]){
      queue.push(neighbor);
    }
  }
  return false;
}

const buildGraph = (edges) => {
  const graph = {}; 
  for(let edge of edges){
    const[a, b] = edge;
    if(!(a in graph)) graph [a] = [];
    if(!(b in graph)) graph [b] = [];
    graph[a].push(b);
    graph[b].push(a);
  }
  return graph;
}

const edges = [
  ['i', 'j'],
  ['k', 'i'],
  ['m', 'k'],
  ['k', 'l'],
  ['o', 'n']
];

undirectedPath(edges, 'j', 'm'); // -> true