DP on Trees Diameter of Tree in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the adjacency list for the tree.

DSA Typescript

const adj: number[][] = Array(n).fill(0).map(() => []);

for (const [u, v] of edges) {
  adj[u].push([1]);
  adj[v].push(u);
}

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

Au

Bv

Cedges

Dadj

Attempts:

3 left

💡 Hint

Common Mistakes

Adding u to u's adjacency list instead of v.

Using the whole edges array instead of a single node.

We add node v to the adjacency list of u to represent the edge u-v.

2fill in blank

medium

Complete the code to perform DFS and update the diameter.

DSA Typescript

function dfs(node: number, parent: number): number {
  let max1 = 0, max2 = 0;
  for (const child of adj[node]) {
    if (child === [1]) continue;
    const depth = dfs(child, node) + 1;
    if (depth > max1) {
      max2 = max1;
      max1 = depth;
    } else if (depth > max2) {
      max2 = depth;
    }
  }
  diameter = Math.max(diameter, max1 + max2);
  return max1;
}

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Aparent

Bnode

Cchild

Ddiameter

Attempts:

3 left

💡 Hint

Common Mistakes

Skipping the wrong node.

Not skipping any node causing infinite recursion.

We skip the parent node to avoid revisiting the node we came from.

3fill in blank

hard

Fix the error in the diameter update line to correctly compute the diameter.

DSA Typescript

diameter = Math.max(diameter, [1] + max2);

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Adepth

Bmax2

Cmax1

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Using only one depth instead of sum of two.

Using wrong variables causing incorrect diameter.

The diameter is updated by the sum of the two longest depths max1 and max2.

4fill in blank

hard

Fill both blanks to complete the function that computes the diameter of the tree.

DSA Typescript

function treeDiameter(n: number, edges: number[][]): number {
  const adj: number[][] = Array(n).fill(0).map(() => []);
  for (const [u, v] of edges) {
    adj[u].push(v);
    adj[v].push(u);
  }
  let diameter = 0;
  function dfs(node: number, parent: number): number {
    let max1 = 0, max2 = 0;
    for (const child of adj[node]) {
      if (child === [1]) continue;
      const depth = dfs(child, node) + 1;
      if (depth > max1) {
        max2 = max1;
        max1 = depth;
      } else if (depth > max2) {
        max2 = depth;
      }
    }
    diameter = Math.max(diameter, max1 + [2]);
    return max1;
  }
  dfs(0, -1);
  return diameter;
}

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Aparent

Bnode

Cmax2

Dmax1

Attempts:

3 left

💡 Hint

Common Mistakes

Not skipping parent causing infinite recursion.

Using wrong variables in diameter update.

Skip the parent node and update diameter with sum of max1 and max2 depths.

5fill in blank

hard

Fill all three blanks to complete the code that calculates the diameter of a tree using DFS.

DSA Typescript

function treeDiameter(n: number, edges: number[][]): number {
  const adj: number[][] = Array(n).fill(0).map(() => []);
  for (const [u, v] of edges) {
    adj[u].push(v);
    adj[v].push(u);
  }
  let diameter = 0;
  function dfs(node: number, parent: number): number {
    let max1 = 0, max2 = 0;
    for (const child of adj[node]) {
      if (child === [1]) continue;
      const depth = dfs(child, [2]) + 1;
      if (depth > max1) {
        max2 = max1;
        max1 = depth;
      } else if (depth > max2) {
        max2 = depth;
      }
    }
    diameter = Math.max(diameter, max1 + max2);
    return [3];
  }
  dfs(0, -1);
  return diameter;
}

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Aparent

Bnode

Cmax1

Dmax2

Attempts:

3 left

💡 Hint

Common Mistakes

Passing wrong parent in recursion.

Returning wrong depth value.

Skip parent node, pass current node as parent in recursion, and return max1 depth.

Practice

1. What does the diameter of a tree represent in graph theory?

easy

A. The longest path between any two nodes in the tree

B. The number of nodes in the tree

C. The height of the tree from the root

D. The number of leaf nodes in the tree

Solution

1. Step 1: Understand the definition of diameter

   The diameter is the longest path between any two nodes in a tree, not just from the root.

2. Step 2: Differentiate from other tree properties

   Height is the longest path from root to leaf, number of nodes and leaves are counts, not paths.

3. Final Answer:

   The longest path between any two nodes in the tree -> Option A

4. Quick Check:

   Diameter = Longest path [OK]

Hint: Diameter is longest path between any two nodes [OK]

Common Mistakes:

• Confusing diameter with height
• Thinking diameter counts nodes instead of path length
• Assuming diameter is from root only

2. Which TypeScript data structure is best to represent a tree for diameter calculation?

easy

A. Boolean flag

B. String array

C. Number variable

D. Array of edges (adjacency list)

Solution

1. Step 1: Identify tree representation needs

   We need to store connections between nodes, so adjacency list (array of edges) is suitable.

2. Step 2: Eliminate unsuitable types

   String array, number variable, or boolean flag cannot represent node connections effectively.

3. Final Answer:

   Array of edges (adjacency list) -> Option D

4. Quick Check:

   Adjacency list = Array of edges [OK]

Hint: Use adjacency list (array of edges) for tree representation [OK]

Common Mistakes:

• Using simple variables instead of lists
• Confusing data types with tree structure
• Using string arrays for numeric node connections

3. What is the output of the following TypeScript code that calculates the diameter of a tree?

const tree = {
  0: [1, 2],
  1: [0, 3],
  2: [0],
  3: [1]
};

function dfs(node: number, parent: number): number {
  let max1 = 0, max2 = 0;
  for (const child of tree[node]) {
    if (child === parent) continue;
    const dist = dfs(child, node) + 1;
    if (dist > max1) {
      max2 = max1;
      max1 = dist;
    } else if (dist > max2) {
      max2 = dist;
    }
  }
  diameter = Math.max(diameter, max1 + max2);
  return max1;
}

let diameter = 0;
dfs(0, -1);
console.log(diameter);

medium

A. 2

B. 3

C. 4

D. 1

Solution

1. Step 1: Trace DFS from node 0

   From node 0, children are 1 and 2. DFS(1) returns 2 (path 1->3), DFS(2) returns 1. dist from 1: 3, from 2: 2.

2. Step 2: Calculate diameter updates

   At node 0, max1=2, max2=1, diameter updated to 3 (2+1). No larger path found elsewhere.

3. Final Answer:

   3 -> Option B

4. Quick Check:

   Longest path length = 3 [OK]

Hint: Sum two longest child paths at each node for diameter [OK]

Common Mistakes:

• Ignoring parent to avoid cycles
• Not updating diameter with sum of two longest paths
• Returning wrong distance from DFS

4. Find the bug in this TypeScript code snippet that tries to compute the diameter of a tree:

const tree = {
  0: [1],
  1: [0, 2],
  2: [1]
};

let diameter = 0;
function dfs(node: number): number {
  let max1 = 0, max2 = 0;
  for (const child of tree[node]) {
    const dist = dfs(child) + 1;
    if (dist > max1) {
      max2 = max1;
      max1 = dist;
    } else if (dist > max2) {
      max2 = dist;
    }
  }
  diameter = Math.max(diameter, max1 + max2);
  return max1;
}
dfs(0);
console.log(diameter);

medium

A. Missing parent parameter to avoid revisiting nodes

B. Diameter variable is not initialized

C. DFS function should return diameter instead of max1

D. Tree is not defined as an array

Solution

1. Step 1: Identify infinite recursion cause

   DFS revisits parent node because no parent check, causing infinite recursion.

2. Step 2: Fix by adding parent parameter

   Pass parent node to DFS and skip it in children loop to avoid cycles.

3. Final Answer:

   Missing parent parameter to avoid revisiting nodes -> Option A

4. Quick Check:

   Parent check prevents infinite recursion [OK]

Hint: Pass parent node in DFS to avoid cycles [OK]

Common Mistakes:

• Not skipping parent node in DFS
• Confusing diameter initialization
• Returning wrong values from DFS

5. Given a weighted tree represented as an adjacency list with edges and weights, how would you modify the diameter calculation to find the longest weighted path?

const tree = {
  0: [{node:1, weight:3}, {node:2, weight:2}],
  1: [{node:0, weight:3}, {node:3, weight:4}],
  2: [{node:0, weight:2}],
  3: [{node:1, weight:4}]
};

hard

A. Ignore weights and count edges only

B. Use BFS instead of DFS without weight consideration

C. Add edge weights to distances during DFS and update diameter accordingly

D. Calculate diameter by counting number of nodes only

Solution

1. Step 1: Incorporate weights in DFS distance calculation

   Instead of adding 1 for each edge, add the edge's weight to the distance.

2. Step 2: Update diameter with sum of two longest weighted paths

   Keep track of two longest weighted distances from children and update diameter with their sum.

3. Final Answer:

   Add edge weights to distances during DFS and update diameter accordingly -> Option C

4. Quick Check:

   Weighted distances sum = diameter [OK]

Hint: Sum edge weights in DFS for weighted diameter [OK]

Common Mistakes:

• Ignoring weights and counting edges only
• Using BFS without weights
• Counting nodes instead of path weights