Shortest Path in Unweighted Graph Using BFS in DSA Typescript

Mental Model

We find the shortest path by exploring neighbors level by level, like ripples spreading in water.

Analogy: Imagine you are in a maze and shout out loud. The sound reaches rooms closest to you first, then rooms further away. The first time the sound reaches a room is the shortest path to that room.

Graph nodes connected:
0 -> 1 -> 3
↓    ↓
2    4

Start at node 0, explore neighbors 1 and 2 first, then their neighbors.

Dry Run Walkthrough

Input: Graph edges: 0-1, 0-2, 1-3, 1-4; Find shortest path from 0 to 4

Goal: Find the shortest path length and the path itself from node 0 to node 4

Step 1: Start BFS from node 0, mark distance 0

Queue: [0]
Distances: {0:0}
Parents: {0:null}

Why: We begin from the start node with distance zero

Step 2: Dequeue 0, enqueue neighbors 1 and 2 with distance 1

Queue: [1, 2]
Distances: {0:0, 1:1, 2:1}
Parents: {0:null, 1:0, 2:0}

Why: Neighbors of 0 are one step away

Step 3: Dequeue 1, enqueue neighbors 3 and 4 with distance 2

Queue: [2, 3, 4]
Distances: {0:0, 1:1, 2:1, 3:2, 4:2}
Parents: {0:null, 1:0, 2:0, 3:1, 4:1}

Why: Neighbors of 1 are two steps away from start

Step 4: Dequeue 2, no new neighbors to enqueue

Queue: [3, 4]
Distances unchanged
Parents unchanged

Why: Node 2 has no unvisited neighbors

Step 5: Dequeue 3, no new neighbors to enqueue

Queue: [4]
Distances unchanged
Parents unchanged

Why: Node 3 has no unvisited neighbors

Step 6: Dequeue 4, target found at distance 2

Queue: []
Distances: {0:0,1:1,2:1,3:2,4:2}
Parents unchanged

Why: Reached target node, shortest path found

Result:

Shortest path length: 2
Path: 0 -> 1 -> 4

Annotated Code

DSA Typescript

class Graph {
  adjacencyList: Map<number, number[]>;

  constructor() {
    this.adjacencyList = new Map();
  }

  addEdge(u: number, v: number) {
    if (!this.adjacencyList.has(u)) this.adjacencyList.set(u, []);
    if (!this.adjacencyList.has(v)) this.adjacencyList.set(v, []);
    this.adjacencyList.get(u)!.push(v);
    this.adjacencyList.get(v)!.push(u);
  }

  shortestPathBFS(start: number, target: number): {distance: number, path: number[]} {
    const queue: number[] = [];
    const distances: Map<number, number> = new Map();
    const parents: Map<number, number | null> = new Map();

    queue.push(start);
    distances.set(start, 0);
    parents.set(start, null);

    while (queue.length > 0) {
      const current = queue.shift()!;
      if (current === target) break;

      for (const neighbor of this.adjacencyList.get(current) || []) {
        if (!distances.has(neighbor)) {
          queue.push(neighbor); // enqueue neighbor
          distances.set(neighbor, distances.get(current)! + 1); // set distance
          parents.set(neighbor, current); // track path
        }
      }
    }

    if (!distances.has(target)) {
      return { distance: -1, path: [] };
    }

    const path: number[] = [];
    let node: number | null = target;
    while (node !== null) {
      path.push(node);
      node = parents.get(node) || null;
    }
    path.reverse();

    return { distance: distances.get(target)!, path };
  }
}

// Driver code
const graph = new Graph();
graph.addEdge(0, 1);
graph.addEdge(0, 2);
graph.addEdge(1, 3);
graph.addEdge(1, 4);

const result = graph.shortestPathBFS(0, 4);
console.log(`Shortest path length: ${result.distance}`);
console.log(`Path: ${result.path.join(' -> ')}`);

queue.push(start); distances.set(start, 0); parents.set(start, null);

Initialize BFS with start node distance zero and no parent

const current = queue.shift()!; if (current === target) break;

Dequeue node and stop if target found

for (const neighbor of this.adjacencyList.get(current) || []) { if (!distances.has(neighbor)) { queue.push(neighbor); distances.set(neighbor, distances.get(current)! + 1); parents.set(neighbor, current); } }

Enqueue unvisited neighbors, set their distance and parent

if (!distances.has(target)) { return { distance: -1, path: [] }; }

Handle case when target is unreachable

while (node !== null) { path.push(node); node = parents.get(node) || null; } path.reverse();

Reconstruct path from target to start using parents

Output

Shortest path length: 2

Path: 0 -> 1 -> 4

Complexity Analysis

Time: O(V + E) because BFS visits every vertex and edge once

Space: O(V) for storing distances, parents, and the queue

vs Alternative: Compared to DFS, BFS guarantees shortest path in unweighted graphs with similar time but DFS may not find shortest path

Edge Cases

Start node equals target node

Distance is zero and path contains only the start node

DSA Typescript

if (current === target) break;

Target node unreachable from start

Returns distance -1 and empty path

DSA Typescript

if (!distances.has(target)) {
  return { distance: -1, path: [] };
}

Graph with isolated nodes

Isolated nodes are never visited, so unreachable targets handled correctly

DSA Typescript

for (const neighbor of this.adjacencyList.get(current) || [])

When to Use This Pattern

When you need the shortest path in a graph without weights, use BFS because it explores nodes in order of distance from the start.

Common Mistakes

Mistake: Not marking nodes as visited before enqueueing causes repeated visits and infinite loops
Fix: Mark nodes as visited (set distance) immediately when enqueueing

Mistake: Using DFS instead of BFS for shortest path in unweighted graph
Fix: Use BFS because it guarantees shortest path by level order traversal

Summary

Finds shortest path in an unweighted graph using breadth-first search.

Use when you want the minimum number of edges between two nodes in an unweighted graph.

BFS explores neighbors level by level, ensuring the first time you reach a node is the shortest path.