DP on Trees Diameter of Tree in DSA Typescript - Execution Trace

Concept Flow - DP on Trees Diameter of Tree

Start at any node

↓

Run DFS to find farthest node A

↓

Run DFS from node A to find farthest node B

↓

Diameter is distance between A and B

↓

Use DP to store max depths during DFS

↓

Update diameter with max sum of two depths at each node

↓

Return final diameter

Find diameter by two DFS traversals and use DP to store max depths for efficient calculation.

Execution Sample

DSA Typescript

function diameterOfTree(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 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;
  }
  dfs(0, -1);
  return diameter;
}

Calculate tree diameter by DFS and DP storing max depths at each node.

Execution Table

Step	Operation	Node Visited	Max Depths (max1, max2)	Diameter Update	Visual State
1	Start DFS from node 0	0	(0,0)	diameter=0	0
2	Visit child 1 of node 0	1	(0,0)	diameter=0	0 -> 1
3	Visit child 3 of node 1	3	(0,0)	diameter=0	0 -> 1 -> 3
4	Return depth 1 from node 3	3	(0,0)	diameter=0	0 -> 1 -> 3
5	Update max depths at node 1: max1=1, max2=0	1	(1,0)	diameter=0	0 -> 1
6	Visit child 2 of node 0	2	(0,0)	diameter=0	0 -> 2
7	Visit child 4 of node 2	4	(0,0)	diameter=0	0 -> 2 -> 4
8	Return depth 1 from node 4	4	(0,0)	diameter=0	0 -> 2 -> 4
9	Update max depths at node 2: max1=1, max2=0	2	(1,0)	diameter=0	0 -> 2
10	Update max depths at node 0: max1=1 (from 1), max2=1 (from 2)	0	(1,1)	diameter=2	0
11	Diameter updated to max1+max2=2	0	(1,1)	diameter=2	Tree diameter path length = 2
12	DFS complete, return diameter	-	-	diameter=2	Final diameter = 2

💡 DFS completes after visiting all nodes; diameter is max sum of two max depths at any node.

Variable Tracker

Variable	Start	After Step 4	After Step 5	After Step 9	After Step 10	Final
diameter	0	0	0	0	2	2
max1 (node 0)	0	0	0	0	1	1
max2 (node 0)	0	0	0	0	1	1
max1 (node 1)	0	1	1	1	1	1
max2 (node 1)	0	0	0	0	0	0
max1 (node 2)	0	0	0	1	1	1
max2 (node 2)	0	0	0	0	0	0

Key Moments - 3 Insights

Why do we keep track of two max depths (max1 and max2) at each node?

Because the diameter can pass through the current node combining two longest paths from its children. See execution_table step 10 where diameter updates using max1 + max2.

Why do we ignore the parent node during DFS?

To avoid revisiting the node we came from and prevent infinite loops. This is shown in the code snippet where child === parent is skipped.

How does the diameter relate to the max depths stored at nodes?

Diameter is the longest path between any two nodes, which equals the sum of two largest depths from a node. Execution_table step 11 shows diameter updated as max1 + max2.

Visual Quiz - 3 Questions

Look at the execution_table at step 10, what are the max depths (max1, max2) at node 0?

A (0,0)

B (2,0)

C (1,1)

D (1,0)

💡 Hint

Check the 'Max Depths (max1, max2)' column at step 10 in execution_table.

At which step does the diameter first update to a non-zero value?

A Step 5

B Step 10

C Step 9

D Step 12

💡 Hint

Look at the 'Diameter Update' column in execution_table to find when diameter changes from 0.

If the tree had only one node, what would be the diameter according to variable_tracker?

A 0

B 1

C Undefined

D 2

💡 Hint

Diameter starts at 0 and only updates when there are edges; see 'diameter' variable in variable_tracker.

Concept Snapshot

DP on Trees Diameter:
- Use DFS twice or DP during DFS
- Track two max depths (max1, max2) at each node
- Diameter = max sum of max1 + max2 over all nodes
- Ignore parent in DFS to avoid cycles
- Return final diameter after DFS completes

Full Transcript

To find the diameter of a tree, we run a depth-first search (DFS) starting from any node to find the farthest node A. Then, we run DFS again from node A to find the farthest node B. The distance between A and B is the diameter. During DFS, we use dynamic programming to store the two largest depths (max1 and max2) from each node's children. The diameter is updated as the maximum sum of these two depths at any node. We skip the parent node during DFS to avoid revisiting nodes. The final diameter is returned after all nodes are visited.

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