55. Find if Path Exists in Graph

class Solution:
    def validPath(self, n: int, edges: List[List[int]], source: int, destination: int) -> bool:
        adj = defaultdict(list)
        for a, b in edges:
            adj[a].append(b)
            adj[b].append(a)
        seen, q = {source}, deque([source])
        while q:                                   # BFS
            u = q.popleft()
            if u == destination:
                return True
            for v in adj[u]:
                if v not in seen:
                    seen.add(v)
                    q.append(v)
        return False

class Solution {
    public boolean validPath(int n, int[][] edges, int source, int destination) {
        List<List<Integer>> adj = new ArrayList<>();
        for (int i = 0; i < n; i++) adj.add(new ArrayList<>());
        for (int[] e : edges) { adj.get(e[0]).add(e[1]); adj.get(e[1]).add(e[0]); }
        boolean[] seen = new boolean[n];
        ArrayDeque<Integer> q = new ArrayDeque<>();
        q.add(source); seen[source] = true;
        while (!q.isEmpty()) {                     // BFS
            int u = q.poll();
            if (u == destination) return true;
            for (int v : adj.get(u)) if (!seen[v]) { seen[v] = true; q.add(v); }
        }
        return false;
    }
}

class Solution {
public:
    bool validPath(int n, vector<vector<int>>& edges, int source, int destination) {
        vector<vector<int>> adj(n);
        for (auto& e : edges) { adj[e[0]].push_back(e[1]); adj[e[1]].push_back(e[0]); }
        vector<bool> seen(n, false);
        queue<int> q;
        q.push(source); seen[source] = true;
        while (!q.empty()) {                       // BFS
            int u = q.front(); q.pop();
            if (u == destination) return true;
            for (int v : adj[u]) if (!seen[v]) { seen[v] = true; q.push(v); }
        }
        return false;
    }
};

function validPath(n, edges, source, destination) {
  const adj = Array.from({ length: n }, () => []);
  for (const [a, b] of edges) { adj[a].push(b); adj[b].push(a); }
  const seen = new Array(n).fill(false), q = [source];
  seen[source] = true;
  for (let h = 0; h < q.length; h++) { // BFS
    const u = q[h];
    if (u === destination) return true;
    for (const v of adj[u]) if (!seen[v]) { seen[v] = true; q.push(v); }
  }
  return false;
}