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.ts

const adj: Map<number, number[]> = new Map([
    [1, [2, 3]],
    [2, [1, 4]],
    [3, [1, 4]],
    [4, [2, 3, 5]],
    [5, [4, 6]],
    [6, [5]],
]);
const src = 1;
const dst = 6;
const dist: Map<number, number> = new Map([[src, 0]]);
const parent: Map<number, number> = new Map([[src, 0]]);
const queue: number[] = [src];
while (queue.length > 0) {
    const v = queue.shift() as number;
    for (const nb of adj.get(v)!) {
        if (!dist.has(nb)) {
            dist.set(nb, (dist.get(v) as number) + 1);
            parent.set(nb, v);
            queue.push(nb);
        }
    }
}
const path: number[] = [];
let node = dst;
while (node !== 0) {
    path.push(node);
    node = parent.get(node) as number;
}
path.reverse();
console.log(JSON.stringify(path));
console.log(dist.get(dst));

Complexity

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

Implementation notes

TypeScript: a dist Map doubles as the visited check, parent records predecessors (0 marks the source), and queue.shift() dequeues.
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.