Shortest Path (Unweighted, via BFS)

BFS explores a graph layer by layer, so the first time it reaches a vertex is along a shortest path. Track dist[v] and parent[v] while exploring, then walk parents back from the target to reconstruct the route.

Algorithm

On the canonical graph from graph-adjacency-list, the shortest path from 1 to 6 is [1, 2, 4, 5, 6] with distance 4. The path is rebuilt from parent: 6 -> 5 -> 4 -> 2 -> 1, reversed.

Basic Implementation

basic.scala

import scala.collection.mutable.{HashMap, Queue, ArrayBuffer}
object Main {
	def main(args: Array[String]): Unit = {
		val adj = HashMap.empty[Int, List[Int]]
		adj(1) = List(2, 3)
		adj(2) = List(1, 4)
		adj(3) = List(1, 4)
		adj(4) = List(2, 3, 5)
		adj(5) = List(4, 6)
		adj(6) = List(5)
		val src = 1
		val dst = 6
		val dist = HashMap.empty[Int, Int]
		val parent = HashMap.empty[Int, Int]
		dist(src) = 0
		parent(src) = 0
		val queue = Queue.empty[Int]
		queue.enqueue(src)
		while (queue.nonEmpty) {
			val v = queue.dequeue()
			for (nb <- adj(v)) {
				if (!dist.contains(nb)) {
					dist(nb) = dist(v) + 1
					parent(nb) = v
					queue.enqueue(nb)
				}
			}
		}
		val path = ArrayBuffer.empty[Int]
		var node = dst
		while (node != 0) {
			path += node
			node = parent(node)
		}
		println(path.reverse.mkString("[", ", ", "]"))
		println(dist(dst))
	}
}

Complexity

Time: O(V + E)
Space: O(V)

Implementation notes

Scala: a dist map doubles as the visited check, parent records predecessors (0 marks the source), and a Queue gives FIFO order.
The replay shows dist, parent, and the queue filling in, then the reconstructed path, matching the lesson spec.

layers equal distance BFS order equals distance in an unweighted graph.