DFS Depth First Search on Graph in DSA Typescript - Execution Trace

Concept Flow - DFS Depth First Search on Graph

Start at root node

↓

Mark node as visited

↓

Explore first unvisited neighbor

↓

Recursive DFS call on neighbor

↓

All neighbors visited?

No→Repeat for next neighbor

Yes↓

Backtrack to previous node

↓

All nodes visited?

No→Continue

Yes↓

End

Start from a node, visit it, then recursively visit each unvisited neighbor deeply before backtracking.

Execution Sample

DSA Typescript

function dfs(node, visited) {
  visited.add(node);
  for (const neighbor of graph[node]) {
    if (!visited.has(neighbor)) dfs(neighbor, visited);
  }
}

This code visits a node, marks it visited, then recursively visits all unvisited neighbors.

Execution Table

Step	Operation	Current Node	Visited Set	Stack (Call Trace)	Action Taken	Graph State
1	Start DFS	A	{}	[]	Visit A	A: [B, C], B: [D], C: [E], D: [], E: []
2	Visit Node	A	{A}	[A]	Mark A visited	Same
3	Explore neighbor	A	{A}	[A]	Check neighbor B	Same
4	Visit Node	B	{A}	[A, B]	Mark B visited	Same
5	Explore neighbor	B	{A, B}	[A, B]	Check neighbor D	Same
6	Visit Node	D	{A, B}	[A, B, D]	Mark D visited	Same
7	Explore neighbor	D	{A, B, D}	[A, B, D]	No neighbors to visit	Same
8	Backtrack	D	{A, B, D}	[A, B]	Return to B	Same
9	All neighbors visited?	B	{A, B, D}	[A, B]	Yes, backtrack	Same
10	Backtrack	B	{A, B, D}	[A]	Return to A	Same
11	Explore neighbor	A	{A, B, D}	[A]	Check neighbor C	Same
12	Visit Node	C	{A, B, D}	[A, C]	Mark C visited	Same
13	Explore neighbor	C	{A, B, D, C}	[A, C]	Check neighbor E	Same
14	Visit Node	E	{A, B, D, C}	[A, C, E]	Mark E visited	Same
15	Explore neighbor	E	{A, B, D, C, E}	[A, C, E]	No neighbors to visit	Same
16	Backtrack	E	{A, B, D, C, E}	[A, C]	Return to C	Same
17	All neighbors visited?	C	{A, B, D, C, E}	[A, C]	Yes, backtrack	Same
18	Backtrack	C	{A, B, D, C, E}	[A]	Return to A	Same
19	All neighbors visited?	A	{A, B, D, C, E}	[A]	Yes, DFS complete	Same
20	End DFS	-	{A, B, D, C, E}	[]	All nodes visited	Same

💡 All nodes visited, no unvisited neighbors remain, DFS ends

Variable Tracker

Variable	Start	After Step 2	After Step 4	After Step 6	After Step 12	After Step 14	Final
visited	{}	{A}	{A, B}	{A, B, D}	{A, B, D, C}	{A, B, D, C, E}	{A, B, D, C, E}
stack	[]	[A]	[A, B]	[A, B, D]	[A, C]	[A, C, E]	[]
current_node	-	A	B	D	C	E	-

Key Moments - 3 Insights

Why do we mark a node as visited before exploring its neighbors?

Marking a node visited early (see Step 2) prevents revisiting it during neighbor checks (Step 3), avoiding infinite loops.

Why do we backtrack after visiting all neighbors?

Backtracking (Steps 8, 10, 16, 18) happens when no unvisited neighbors remain, so DFS returns to previous nodes to explore other paths.

Why does the stack grow and shrink during DFS?

The stack (call trace) grows when DFS goes deeper (Steps 4, 6, 12, 14) and shrinks when backtracking (Steps 8, 10, 16, 18), reflecting recursion depth.

Visual Quiz - 3 Questions

Look at the execution table, what is the visited set after Step 6?

A {A, B}

B {A, B, D}

C {A, B, C}

D {A, B, D, C}

💡 Hint

Check the 'Visited Set' column at Step 6 in the execution table.

At which step does DFS first backtrack from node D?

A Step 8

B Step 7

C Step 9

D Step 10

💡 Hint

Look for 'Backtrack' operation with current node D in the execution table.

If node E had a neighbor F not visited yet, what would happen at Step 15?

A DFS would skip F and finish

B DFS would backtrack immediately

C DFS would visit F next, stack grows

D DFS would restart from A

💡 Hint

At Step 15, DFS explores neighbors; unvisited neighbors cause recursive calls and stack growth.

Concept Snapshot

DFS visits nodes deeply before backtracking.
Start at a node, mark visited.
Recursively visit unvisited neighbors.
Use stack (call trace) to track path.
Backtrack when no neighbors left.
Ends when all reachable nodes visited.

Full Transcript

Depth First Search (DFS) starts at a chosen node, marks it visited, then explores each neighbor deeply before backtracking. The process uses recursion, which acts like a stack to remember the path. Each step marks nodes visited to avoid repeats. When a node has no unvisited neighbors, DFS backtracks to explore other paths. This continues until all nodes reachable from the start are visited. The execution table shows each step, the current node, visited nodes, and the call stack. Key moments include marking visited early to prevent loops, backtracking when no neighbors remain, and stack changes reflecting recursion depth.

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