DP on Trees Diameter of Tree in DSA Typescript - Time & Space Complexity

Time Complexity: DP on Trees Diameter of Tree

O(n)

Understanding Time Complexity

We want to understand how long it takes to find the longest path in a tree using dynamic programming.

How does the time grow as the tree gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code finds the longest path (diameter) in a tree using a depth-first search and dynamic programming.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The recursive depth-first search (dfs) visits each node once.
• How many times: Each node and its edges are processed exactly once during the recursion.

How Execution Grows With Input

As the number of nodes grows, the function visits each node and edge once.

Input Size (n)	Approx. Operations
10	About 10 nodes and their edges visited once each
100	About 100 nodes and their edges visited once each
1000	About 1000 nodes and their edges visited once each

Pattern observation: The work grows roughly in direct proportion to the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the diameter grows linearly with the number of nodes in the tree.

Common Mistake

[X] Wrong: "The diameter calculation takes quadratic time because it checks all pairs of nodes."

[OK] Correct: The code uses a single depth-first search that visits each node once, avoiding repeated checks of node pairs.

Interview Connect

Understanding this linear time approach shows you can handle tree problems efficiently, a key skill in many coding challenges.

Self-Check

"What if the tree was represented as an adjacency matrix instead of adjacency lists? How would the time complexity change?"

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