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.swift

var adj: [Int: [Int]] = [:]
adj[1] = [2, 3]
adj[2] = [1, 4]
adj[3] = [1, 4]
adj[4] = [2, 3, 5]
adj[5] = [4, 6]
adj[6] = [5]
let src = 1
let dst = 6
var dist: [Int: Int] = [src: 0]
var parent: [Int: Int] = [src: 0]
var queue: [Int] = [src]
while !queue.isEmpty {
	let v = queue.removeFirst()
	for nb in adj[v]! {
		if dist[nb] == nil {
			dist[nb] = dist[v]! + 1
			parent[nb] = v
			queue.append(nb)
		}
	}
}
var path: [Int] = []
var node = dst
while node != 0 {
	path.append(node)
	node = parent[node]!
}
print(Array(path.reversed()))
print(dist[dst]!)

Complexity

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

Implementation notes

Swift: a dist dictionary doubles as the visited check, parent records predecessors (0 marks the source), and an array index walks the queue.
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.