DFS Depth First Search on Graph in DSA Typescript

Mental Model

DFS explores as far as possible along each branch before backtracking. It dives deep into one path before trying others.

Analogy: Imagine exploring a maze by always taking the first unexplored path you see, going deep until you hit a dead end, then backtracking to try other paths.

Graph adjacency list:
0 -> [1, 2]
1 -> [2]
2 -> [0, 3]
3 -> [3]
Start at node 2 ↑

Dry Run Walkthrough

Input: Graph with edges: 0->1, 0->2, 1->2, 2->0, 2->3, 3->3; start DFS at node 2

Goal: Visit all nodes reachable from node 2 using DFS order

Step 1: Start DFS at node 2, mark 2 visited

Visited: {2}
Stack: [2]

Why: We begin exploring from node 2

Step 2: From 2, go to neighbor 0, mark 0 visited

Visited: {2, 0}
Stack: [2, 0]

Why: DFS goes deep to first unvisited neighbor

Step 3: From 0, go to neighbor 1, mark 1 visited

Visited: {2, 0, 1}
Stack: [2, 0, 1]

Why: Continue deep exploration from 0 to 1

Step 4: From 1, neighbor 2 already visited, backtrack to 0

Visited: {2, 0, 1}
Stack: [2, 0]

Why: No new nodes from 1, so backtrack

Step 5: From 0, neighbor 2 visited, backtrack to 2

Visited: {2, 0, 1}
Stack: [2]

Why: All neighbors of 0 visited, backtrack

Step 6: From 2, go to neighbor 3, mark 3 visited

Visited: {2, 0, 1, 3}
Stack: [2, 3]

Why: Explore next neighbor of 2

Step 7: From 3, neighbor 3 already visited (self-loop), backtrack to 2

Visited: {2, 0, 1, 3}
Stack: [2]

Why: No new nodes from 3, backtrack

Step 8: All neighbors of 2 visited, DFS complete

Visited: {2, 0, 1, 3}
Stack: []

Why: Traversal finished, all reachable nodes visited

Result:

Visited nodes in DFS order: 2 -> 0 -> 1 -> 3

Annotated Code

DSA Typescript

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

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

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

  dfs(start: number): number[] {
    const visited = new Set<number>();
    const result: number[] = [];

    const dfsHelper = (node: number) => {
      visited.add(node); // mark node visited
      result.push(node); // record visit order

      const neighbors = this.adjacencyList.get(node) || [];
      for (const neighbor of neighbors) {
        if (!visited.has(neighbor)) {
          dfsHelper(neighbor); // recursively visit neighbors
        }
      }
    };

    dfsHelper(start);
    return result;
  }
}

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

const dfsOrder = graph.dfs(2);
console.log(dfsOrder.join(' -> '));

visited.add(node); // mark node visited

mark current node as visited to avoid revisiting

result.push(node); // record visit order

record the order of node visits

for (const neighbor of neighbors) { if (!visited.has(neighbor)) { dfsHelper(neighbor); } }

recursively visit each unvisited neighbor to explore deeply

Output

2 -> 0 -> 1 -> 3

Complexity Analysis

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

Space: O(V) for the visited set and recursion stack in worst case

vs Alternative: Compared to BFS which uses a queue and explores level by level, DFS uses recursion and explores depth first; both have similar time but different traversal orders

Edge Cases

Graph with no edges (isolated nodes)

DFS visits only the start node since no neighbors exist

DSA Typescript

const neighbors = this.adjacencyList.get(node) || [];

Graph with self-loops

Self-loops do not cause infinite loops because visited nodes are tracked

DSA Typescript

if (!visited.has(neighbor)) { dfsHelper(neighbor); }

Start node not in graph

DFS visits only the start node as it has no neighbors

DSA Typescript

const neighbors = this.adjacencyList.get(node) || [];

When to Use This Pattern

When a problem requires exploring all connected parts of a graph deeply before backtracking, use DFS to traverse paths fully before trying alternatives.

Common Mistakes

Mistake: Not marking nodes as visited before recursive calls, causing infinite loops on cycles
Fix: Mark nodes as visited immediately when first visiting them to prevent revisiting

Summary

DFS visits nodes by going deep along each path before backtracking.

Use DFS when you want to explore all nodes reachable from a start point deeply.

The key is to mark nodes visited as soon as you see them to avoid cycles.

Practice

1. What is the main idea behind Depth First Search (DFS) on a graph?

easy

A. Explore as far as possible along each branch before backtracking

B. Visit all neighbors of a node before moving deeper

C. Visit nodes in order of their distance from the start node

D. Randomly visit nodes until all are visited

Solution

1. Step 1: Understand DFS traversal strategy

   DFS explores nodes by going deep into one branch before backtracking to explore other branches.

2. Step 2: Compare with other traversal methods

   BFS visits neighbors first, but DFS goes deep first, matching Explore as far as possible along each branch before backtracking.

3. Final Answer:

   Explore as far as possible along each branch before backtracking -> Option A

4. Quick Check:

   DFS = deep exploration first [OK]

Hint: DFS goes deep first, BFS goes wide first [OK]

Common Mistakes:

• Confusing DFS with BFS
• Thinking DFS visits neighbors first
• Assuming DFS visits nodes by distance

2. Which of the following is the correct TypeScript function signature for a DFS function on a graph represented as an adjacency list?

easy

A. function dfs(graph: Map, start: number): void

B. function dfs(graph: number[], start: number): void

C. function dfs(graph: number[][], start: number): void

D. function dfs(graph: string[][], start: string): void

Solution

1. Step 1: Identify graph representation type

   An adjacency list for a graph with numbered nodes is usually number[][], where each index holds neighbors.

2. Step 2: Check parameter types

   Start node is a number, so function signature with graph: number[][] and start: number is correct.

3. Final Answer:

   function dfs(graph: number[][], start: number): void -> Option C

4. Quick Check:

   Adjacency list = number[][], start node = number [OK]

Hint: Adjacency list uses array of arrays for neighbors [OK]

Common Mistakes:

• Using number[] instead of number[][] for adjacency list
• Using Map with incorrect value type
• Using string types for numeric graphs

3. What is the output of the following DFS traversal starting from node 0?

const graph = [[1, 2], [2], [0, 3], [3]];
const visited: boolean[] = [];
const result: number[] = [];
function dfs(node: number) {
  if (visited[node]) return;
  visited[node] = true;
  result.push(node);
  for (const neighbor of graph[node]) {
    dfs(neighbor);
  }
}
dfs(0);
console.log(result);

medium

A. [0, 2, 3, 1]

B. [0, 2, 1, 3]

C. [0, 1, 2]

D. [0, 1, 2, 3]

Solution

1. Step 1: Start DFS at node 0

   Visit 0, mark visited, add to result: [0]. Neighbors: 1, 2.

2. Step 2: Visit neighbor 1

   Visit 1, mark visited, add: [0,1]. Neighbor: 2.

3. Step 3: Visit neighbor 2 from 1

   Visit 2, mark visited, add: [0,1,2]. Neighbors: 0 (visited), 3.

4. Step 4: Visit neighbor 3 from 2

   Visit 3, mark visited, add: [0,1,2,3]. Neighbor: 3 (already visited).

5. Final Answer:

   [0, 2, 1, 3] -> Option B

6. Quick Check:

   DFS order = [0,2,1,3] [OK]

Hint: Follow neighbors recursively, mark visited to avoid repeats [OK]

Common Mistakes:

• Not marking nodes visited before recursion
• Visiting neighbors in wrong order
• Ignoring already visited nodes

4. Identify the bug in this DFS implementation and choose the fix:

function dfs(node: number, graph: number[][], visited: boolean[]) {
  visited[node] = true;
  for (const neighbor of graph[node]) {
    if (!visited[neighbor]) {
      dfs(neighbor, graph, visited);
    }
  }
}

const graph = [[1], [2], [0]];
const visited = new Array(graph.length).fill(false);
dfs(0, graph, visited);
console.log(visited);

medium

A. Mark visited[node] = true after recursive calls

B. Initialize visited array with true instead of false

C. Add a base case to return if node is already visited at start

D. No bug, code works correctly

Solution

1. Step 1: Check visited marking timing

   Visited[node] is marked true before recursion, preventing revisits.

2. Step 2: Verify recursion and base case

   Recursion only happens if neighbor not visited, so no infinite loops.

3. Final Answer:

   No bug, code works correctly -> Option D

4. Quick Check:

   Visited marked before recursion = correct DFS [OK]

Hint: Mark visited before recursion to avoid cycles [OK]

Common Mistakes:

• Marking visited after recursion causing infinite loops
• Initializing visited array incorrectly
• Missing base case check

5. Given a graph with possible cycles, how can you modify DFS to detect if a cycle exists starting from a given node? Choose the best approach.

hard

A. Use a recursion stack array to track nodes in the current path and detect back edges

B. Only mark nodes visited after all neighbors are visited

C. Use BFS instead of DFS to detect cycles

D. Ignore visited nodes and continue traversal

Solution

1. Step 1: Understand cycle detection in DFS

   Detecting cycles requires knowing if a node is in the current recursion path (stack).

2. Step 2: Use recursion stack tracking

   Maintain a separate array to track nodes currently in recursion; if a neighbor is in this stack, a cycle exists.

3. Step 3: Evaluate other options

   Marking visited after neighbors or using BFS won't reliably detect cycles; ignoring visited nodes causes infinite loops.

4. Final Answer:

   Use a recursion stack array to track nodes in the current path and detect back edges -> Option A

5. Quick Check:

   Cycle detection = recursion stack tracking [OK]

Hint: Track recursion path to find back edges indicating cycles [OK]

Common Mistakes:

• Not tracking recursion stack causing missed cycles
• Using BFS which doesn't track recursion path
• Marking visited too late or ignoring visited nodes