Graph vs Tree Key Structural Difference in DSA Javascript - Interactive Comparison Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if a graph has cycles.

DSA Javascript

function hasCycle(graph) {
  const visited = new Set();
  function dfs(node, parent) {
    visited.add(node);
    for (const neighbor of graph[node]) {
      if (!visited.has(neighbor)) {
        if (dfs(neighbor, node)) return true;
      } else if (neighbor !== [1]) {
        return true;
      }
    }
    return false;
  }
  for (const node in graph) {
    if (!visited.has(node)) {
      if (dfs(node, null)) return true;
    }
  }
  return false;
}

Drag options to blanks, or click blank then click option'

Anode

Bgraph

Cparent

Dneighbor

Attempts:

3 left

💡 Hint

Common Mistakes

Using the current node instead of the parent node causes incorrect cycle detection.

Comparing with the neighbor itself leads to wrong results.

The cycle check compares the neighbor with the parent node to avoid false positives.

2fill in blank

medium

Complete the code to add an edge in an undirected graph.

DSA Javascript

function addEdge(graph, u, v) {
  if (!graph[u]) graph[u] = [];
  if (!graph[v]) graph[v] = [];
  graph[u].push(v);
  graph[v].[1](u);
}

Drag options to blanks, or click blank then click option'

Aappend

Badd

Cinsert

Dpush

Attempts:

3 left

💡 Hint

Common Mistakes

Using append which is not valid in JavaScript arrays.

Using add which is for sets, not arrays.

Use push to add an element to the array representing adjacency.

3fill in blank

hard

Fix the error in the code to check if a tree is a valid tree (no cycles and connected).

DSA Javascript

function isValidTree(n, edges) {
  const graph = {};
  for (const [u, v] of edges) {
    if (!graph[u]) graph[u] = [];
    if (!graph[v]) graph[v] = [];
    graph[u].push(v);
    graph[v].push(u);
  }
  for(let i = 0; i < n; i++) {
    if (!graph[i]) graph[i] = [];
  }
  const visited = new Set();
  function dfs(node, parent) {
    visited.add(node);
    for (const neighbor of graph[node]) {
      if (!visited.has(neighbor)) {
        if (dfs(neighbor, node)) return true;
      } else if (neighbor !== [1]) {
        return true;
      }
    }
    return false;
  }
  if (dfs(0, -1)) return false;
  return visited.size === n;
}

Drag options to blanks, or click blank then click option'

Aparent

Bnode

Cneighbor

Dgraph

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with the current node instead of the parent.

Not excluding the parent node causes false cycle detection.

The cycle detection must ignore the parent node to avoid false positives.

4fill in blank

hard

Complete the code to create an object that maps words to their lengths only if length is greater than 3.

DSA Javascript

const words = ['tree', 'graph', 'node', 'edge'];
const result = {};
for (const word of words) {
  if (word.length > 3) {
    result[[1]] = [2];
  }
}

Drag options to blanks, or click blank then click option'

Aword

Bword.length

Cword.toUpperCase()

Dword.length > 3

Attempts:

3 left

💡 Hint

Common Mistakes

Using a condition as a key.

Using the word itself as the value instead of its length.

The key is the word itself, and the value is its length.

5fill in blank

hard

Fill in the blanks to create a filtered object of uppercase keys and their lengths for words longer than 4 characters.

DSA Javascript

const words = ['tree', 'graph', 'node', 'edge', 'vertex'];
const filtered = {};
for (const word of words) {
  if (word.length [3] 4) {
    filtered[[1]] = [2];
  }
}

Drag options to blanks, or click blank then click option'

Aword.toUpperCase()

Bword.length

C>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong comparison operator.

Not converting keys to uppercase.

Keys are uppercase words, values are lengths, filtered by length greater than 4.

Practice

1. Which of the following is a key structural difference between a graph and a tree?

easy

A. A tree has no cycles, but a graph can have cycles.

B. A graph always has a root node, but a tree does not.

C. A tree can have multiple edges between the same nodes, but a graph cannot.

D. A graph must be connected, but a tree can be disconnected.

Solution

1. Step 1: Understand tree structure

   A tree is a special type of graph that is connected and has no cycles.

2. Step 2: Compare with graph properties

   Graphs can have cycles and may or may not be connected.

3. Final Answer:

   A tree has no cycles, but a graph can have cycles. -> Option A

4. Quick Check:

   No cycles in tree = key difference [OK]

Hint: Trees never loop back; graphs can have loops [OK]

Common Mistakes:

• Thinking graphs always have a root node
• Believing trees can have cycles
• Confusing connectivity rules between graphs and trees

2. Which syntax correctly represents a tree structure in JavaScript using adjacency lists?

easy

A. const tree = [ [1, 2], [2, 3], [3, 4] ];

B. const tree = { 1: 2, 2: 3, 3: 4 };

C. const tree = { 1: [2, 3], 2: [4], 3: [], 4: [] };

D. const tree = [1, 2, 3, 4];

Solution

1. Step 1: Identify adjacency list format

   An adjacency list uses an object with keys as nodes and values as arrays of connected nodes.

2. Step 2: Check each option

   const tree = { 1: [2, 3], 2: [4], 3: [], 4: [] }; matches adjacency list format for a tree; others do not represent adjacency lists properly.

3. Final Answer:

   const tree = { 1: [2, 3], 2: [4], 3: [], 4: [] }; -> Option C

4. Quick Check:

   Adjacency list uses object with arrays [OK]

Hint: Adjacency list = object keys with array values [OK]

Common Mistakes:

• Using arrays of pairs instead of adjacency lists
• Assigning single values instead of arrays for neighbors
• Confusing array of nodes with adjacency list

3. What will be the output of the following code snippet representing a graph with a cycle?

const graph = { 1: [2], 2: [3], 3: [1] };
function hasCycle(node, visited, stack) {
  if (!visited[node]) {
    visited[node] = true;
    stack[node] = true;
    for (const neighbor of graph[node]) {
      if (!visited[neighbor] && hasCycle(neighbor, visited, stack)) {
        return true;
      } else if (stack[neighbor]) {
        return true;
      }
    }
  }
  stack[node] = false;
  return false;
}
console.log(hasCycle(1, {}, {}));

medium

A. false

B. true

C. undefined

D. TypeError

Solution

1. Step 1: Understand the graph structure

   The graph has edges 1->2, 2->3, and 3->1, forming a cycle.

2. Step 2: Trace the hasCycle function

   Starting at node 1, the function detects the cycle when it revisits node 1 in the recursion stack.

3. Final Answer:

   true -> Option B

4. Quick Check:

   Cycle detected = true [OK]

Hint: Cycle detection returns true if node revisited in recursion [OK]

Common Mistakes:

• Assuming no cycle because of simple code
• Confusing visited and recursion stack usage
• Expecting false without tracing recursion

4. Find the error in the following code that tries to check if a tree has cycles:

const tree = { 1: [2], 2: [3], 3: [] };
function hasCycle(node, visited) {
  if (visited[node]) return true;
  visited[node] = true;
  for (const neighbor of tree[node]) {
    if (hasCycle(neighbor, visited)) return true;
  }
  return false;
}
console.log(hasCycle(1, {}));

medium

A. The function does not track recursion stack, so false positives occur.

B. The visited object is not initialized properly.

C. The tree object keys should be numbers, not strings.

D. The function should return false immediately after visiting a node.

Solution

1. Step 1: Analyze cycle detection logic

   The function uses only visited to detect cycles but misses recursion stack tracking.

2. Step 2: Explain why this causes errors

   Without recursion stack, revisiting a node from a different branch may cause false cycle detection.

3. Final Answer:

   The function does not track recursion stack, so false positives occur. -> Option A

4. Quick Check:

   Missing recursion stack causes wrong cycle detection [OK]

Hint: Track recursion stack to avoid false cycle detection [OK]

Common Mistakes:

• Ignoring recursion stack importance
• Assuming visited alone detects cycles correctly
• Confusing object key types

5. Given a graph represented as adjacency lists, how can you verify if it is a tree? Choose the best approach.

hard

A. Check if the graph has cycles and is disconnected.

B. Check if the graph has no cycles regardless of connectivity.

C. Check if the graph has a root node and all nodes have at most two children.

D. Check if the graph is connected and has exactly n-1 edges for n nodes.

Solution

1. Step 1: Recall tree properties

   A tree with n nodes must be connected and have exactly n-1 edges, with no cycles.

2. Step 2: Analyze options

   Check if the graph is connected and has exactly n-1 edges for n nodes. correctly combines connectivity and edge count to verify a tree. Others miss key conditions.

3. Final Answer:

   Check if the graph is connected and has exactly n-1 edges for n nodes. -> Option D

4. Quick Check:

   Connectivity + edges = tree check [OK]

Hint: Tree = connected + edges = nodes minus one [OK]

Common Mistakes:

• Ignoring connectivity requirement
• Only checking for cycles without edges count
• Assuming root or children count defines tree