DP on Trees Diameter of Tree in DSA Typescript

Mental Model

The diameter of a tree is the longest path between any two nodes. We find it by exploring the longest paths from each node using dynamic programming.

Analogy: Imagine a network of roads connecting cities (nodes). The diameter is the longest route you can drive without repeating roads. To find it, you check the longest routes starting from each city and combine them.

    1
   / \
  2   3
 /     \
4       5

Nodes connected like a tree with branches.

Dry Run Walkthrough

Input: Tree edges: 1-2, 1-3, 2-4, 3-5

Goal: Find the longest path between any two nodes in the tree (diameter).

Step 1: Start DFS from node 1, calculate longest path down each subtree

At node 4: longest path down is 0 (leaf)
At node 5: longest path down is 0 (leaf)

Why: Leaf nodes have no children, so longest downward path is 0

Step 2: At node 2, longest downward path is max of child (node 4) + 1 = 1

Node 2 longest downward path = 1
Node 4 longest downward path = 0

Why: Add 1 for edge from node 2 to node 4

Step 3: At node 3, longest downward path is max of child (node 5) + 1 = 1

Node 3 longest downward path = 1
Node 5 longest downward path = 0

Why: Add 1 for edge from node 3 to node 5

Step 4: At node 1, longest downward paths from children are 1 (node 2) and 1 (node 3)

Node 1 children longest downward paths = [1, 1]

Why: We need these to compute diameter passing through node 1

Step 5: Calculate diameter at node 1 as sum of top two longest downward paths = 1 + 1 = 2

Diameter candidate at node 1 = 2

Why: Longest path passes through node 1 connecting node 4 and node 5

Step 6: Compare diameter candidates from all nodes, max is 2

Final diameter = 2

Why: Diameter is the longest path found in the tree

Result:

2

Annotated Code

DSA Typescript

class Tree {
  adjacencyList: Map<number, number[]> = new Map();

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

  diameter(): number {
    let diameter = 0;
    const visited = new Set<number>();

    const dfs = (node: number): number => {
      visited.add(node);
      let max1 = 0, max2 = 0;

      for (const neighbor of this.adjacencyList.get(node) || []) {
        if (!visited.has(neighbor)) {
          const length = dfs(neighbor) + 1;
          if (length > max1) {
            max2 = max1;
            max1 = length;
          } else if (length > max2) {
            max2 = length;
          }
        }
      }

      diameter = Math.max(diameter, max1 + max2);
      return max1;
    };

    dfs(1);
    return diameter;
  }
}

// Driver code
const tree = new Tree();
tree.addEdge(1, 2);
tree.addEdge(1, 3);
tree.addEdge(2, 4);
tree.addEdge(3, 5);
console.log(tree.diameter());

const dfs = (node: number): number => {

start DFS to explore longest downward paths from node

visited.add(node);

mark node as visited to avoid cycles

for (const neighbor of this.adjacencyList.get(node) || []) {

iterate over neighbors to explore subtrees

if (!visited.has(neighbor)) {

only visit unvisited neighbors to prevent revisiting

const length = dfs(neighbor) + 1;

compute longest path length from neighbor plus edge

if (length > max1) { max2 = max1; max1 = length; } else if (length > max2) { max2 = length; }

track top two longest downward paths from children

diameter = Math.max(diameter, max1 + max2);

update diameter with sum of two longest paths through current node

return max1;

return longest downward path for parent's calculation

Output

2

Complexity Analysis

Time: O(n) because we visit each node once during DFS

Space: O(n) for recursion stack and adjacency list storage

vs Alternative: Naive approach checking all pairs is O(n^2), DP on trees reduces it to O(n)

Edge Cases

Single node tree

Diameter is 0 because no edges exist

DSA Typescript

visited.add(node);

Linear tree (chain)

Diameter equals number of edges in chain

DSA Typescript

diameter = Math.max(diameter, max1 + max2);

Tree with multiple branches of same length

Diameter correctly picks longest path through two longest branches

DSA Typescript

if (length > max1) { max2 = max1; max1 = length; } else if (length > max2) { max2 = length; }

When to Use This Pattern

When asked to find longest path in a tree, use DP on trees by DFS to find longest downward paths and combine them for diameter.

Common Mistakes

Mistake: Not tracking two longest paths from children, only one
Fix: Track top two longest downward paths to compute diameter passing through node

Summary

Finds the longest path between any two nodes in a tree using DFS and dynamic programming.

Use when you need the maximum distance between nodes in a tree structure.

The diameter is the sum of the two longest downward paths from a node.

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