DP on Trees Diameter of Tree in DSA Rust - Build from Scratch

DP on Trees: Diameter of Tree

📖 Scenario: You are working with a network of connected points represented as a tree. Each point is connected by edges without any cycles. You want to find the longest path between any two points in this network, called the diameter of the tree.

🎯 Goal: Build a Rust program that uses dynamic programming on trees to find the diameter (longest path) of a given tree represented by edges.

📋 What You'll Learn

Create a tree using adjacency lists from given edges

Use a helper variable to track the maximum diameter found

Implement a depth-first search (DFS) function to compute the longest paths

Print the diameter of the tree after computation

💡 Why This Matters

🌍 Real World

Finding the diameter of a tree is useful in network design, biology (e.g., evolutionary trees), and computer graphics to understand the longest distance between nodes.

💼 Career

Understanding tree algorithms and dynamic programming is essential for software engineers working on graph problems, optimization, and system design.

1

Create the tree adjacency list

Create a variable called edges as a vector of tuples with these exact pairs: (0, 1), (1, 2), (1, 3), (3, 4). Then create a variable called n and set it to 5 representing the number of nodes. Finally, create a vector of vectors called tree with n empty vectors to hold adjacency lists.

DSA Rust

# Your code here

Hint

Use vec![] to create empty adjacency lists for each node.

2

Build the adjacency list from edges

Add a for loop iterating over edges with variables u and v. Inside the loop, add v to tree[u] and u to tree[v] to build the undirected tree. Also, create a mutable variable called max_diameter and set it to 0 to track the longest path found.

DSA Rust

let edges = vec![(0, 1), (1, 2), (1, 3), (3, 4)];
let n = 5;
let mut tree: Vec> = vec![vec![]; n];
# Your code here

Hint

Remember to add edges both ways because the tree is undirected.

3

Implement DFS to find diameter

Write a recursive function called dfs that takes parameters node: usize, parent: usize, tree: &Vec>, and max_diameter: &mut usize. Inside, create a vector depths to store depths of children. For each child in tree[node] not equal to parent, recursively call dfs and collect depths. Update max_diameter with the sum of the two largest depths if greater. Return the maximum depth plus one.

DSA Rust

let edges = vec![(0, 1), (1, 2), (1, 3), (3, 4)];
let n = 5;
let mut tree: Vec> = vec![vec![]; n];
for (u, v) in edges {
    tree[u].push(v);
    tree[v].push(u);
}
let mut max_diameter = 0;

// Define dfs function here
# Your code here

Hint

Use recursion to explore children and keep track of the two longest depths to update the diameter.

4

Print the diameter of the tree

Write a println! statement to print the value of max_diameter.

DSA Rust

let edges = vec![(0, 1), (1, 2), (1, 3), (3, 4)];
let n = 5;
let mut tree: Vec> = vec![vec![]; n];
for (u, v) in edges {
    tree[u].push(v);
    tree[v].push(u);
}
let mut max_diameter = 0;

fn dfs(node: usize, parent: usize, tree: &Vec>, max_diameter: &mut usize) -> usize {
    let mut depths = Vec::new();
    for &child in &tree[node] {
        if child != parent {
            depths.push(dfs(child, node, tree, max_diameter));
        }
    }
    depths.sort_unstable_by(|a, b| b.cmp(a));
    if depths.len() >= 2 {
        *max_diameter = (*max_diameter).max(depths[0] + depths[1]);
    } else if depths.len() == 1 {
        *max_diameter = (*max_diameter).max(depths[0]);
    }
    if depths.is_empty() {
        1
    } else {
        depths[0] + 1
    }
}

dfs(0, usize::MAX, &tree, &mut max_diameter);

# Your code here

Hint

The diameter of this tree is 4 edges long, so printing max_diameter should output 4.

Practice

1. What does the diameter of a tree represent in the context of DP on Trees?

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 sum of all edge weights in the tree

Solution

1. Step 1: Understand the definition of tree diameter

   The diameter is the longest path between any two nodes, not just from the root or sum of edges.

2. Step 2: Differentiate from other tree properties

   Height is from root to leaf, sum of edges is total weight, number of nodes is size, none equal diameter.

3. Final Answer:

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

4. Quick Check:

   Diameter = Longest path [OK]

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

Common Mistakes:

• Confusing diameter with tree height
• Thinking diameter is total nodes count
• Assuming diameter is sum of all edges

2. Which Rust syntax correctly declares a recursive function to compute the diameter of a tree represented as adjacency list Vec?

easy

A. fn diameter(node: usize, parent: usize, graph: &mut Vec) -> usize { ... }

B. fn diameter(node: usize, graph: Vec) -> usize { ... }

C. fn diameter(node: usize, parent: usize, graph: &Vec) -> usize { ... }

D. fn diameter(&self, node: usize, parent: usize) -> usize { ... }

Solution

1. Step 1: Identify correct parameter passing for graph

   Graph should be passed by reference (&Vec) to avoid copying large data.

2. Step 2: Include parent parameter to avoid revisiting parent node

   Parent node is needed to prevent cycles in undirected tree traversal.

3. Final Answer:

   fn diameter(node: usize, parent: usize, graph: &Vec) -> usize { ... } -> Option C

4. Quick Check:

   Pass graph by reference and track parent [OK]

Hint: Pass graph as &Vec and track parent node [OK]

Common Mistakes:

• Passing graph by value causing performance issues
• Missing parent parameter causing infinite recursion
• Using &mut when mutation is not needed

3. Given the following Rust code snippet to compute the diameter of a tree, what is the printed output?

fn dfs(node: usize, parent: usize, graph: &Vec, max_diameter: &mut usize) -> usize {
    let mut max1 = 0;
    let mut max2 = 0;
    for &child in &graph[node] {
        if child == parent { continue; }
        let height = dfs(child, node, graph, max_diameter);
        if height > max1 {
            max2 = max1;
            max1 = height;
        } else if height > max2 {
            max2 = height;
        }
    }
    *max_diameter = (*max_diameter).max(max1 + max2);
    max1 + 1
}

fn main() {
    let graph = vec![
        vec![1, 2],
        vec![0, 3, 4],
        vec![0],
        vec![1],
        vec![1]
    ];
    let mut max_diameter = 0;
    dfs(0, usize::MAX, &graph, &mut max_diameter);
    println!("{}", max_diameter);
}

medium

A. 3

B. 4

C. 2

D. 5

Solution

1. Step 1: Trace DFS heights from leaves

   Leaves 2, 3, 4 return height 1; node 1 gets max1=1, max2=1; node 0 gets max1=2 (from node 1), max2=1 (from node 2).

2. Step 2: Calculate max diameter updates

   At node 1, max_diameter = max(0,1+1)=2; at node 0, max_diameter = max(2,2+1)=3.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Longest path length = 3 edges [OK]

Hint: Sum two largest subtree heights at each node [OK]

Common Mistakes:

• Counting nodes instead of edges for diameter
• Not updating max_diameter correctly
• Ignoring parent to avoid revisiting

4. Identify the bug in this Rust function computing tree diameter and select the fix:

fn dfs(node: usize, graph: &Vec, max_diameter: &mut usize) -> usize {
    let mut max1 = 0;
    let mut max2 = 0;
    for &child in &graph[node] {
        let height = dfs(child, graph, max_diameter);
        if height > max1 {
            max2 = max1;
            max1 = height;
        } else if height > max2 {
            max2 = height;
        }
    }
    *max_diameter = (*max_diameter).max(max1 + max2);
    max1 + 1
}

medium

A. Initialize max1 and max2 to 1 instead of 0

B. Add a parent parameter to avoid revisiting the parent node

C. Change max_diameter to return instead of using mutable reference

D. Remove max2 variable as it is unnecessary

Solution

1. Step 1: Identify infinite recursion cause

   Without parent tracking, dfs revisits parent causing infinite recursion or stack overflow.

2. Step 2: Fix by adding parent parameter

   Pass parent node to skip revisiting it in children loop, preventing cycles.

3. Final Answer:

   Add a parent parameter to avoid revisiting the parent node -> Option B

4. Quick Check:

   Parent tracking prevents infinite recursion [OK]

Hint: Always track parent node in undirected tree DFS [OK]

Common Mistakes:

• Not passing parent causing infinite recursion
• Misusing mutable references for max_diameter
• Incorrect initialization of max heights

5. You want to find the diameter of a weighted tree where edges have weights. Which modification to the DP on Trees diameter algorithm is correct?

hard

A. Ignore weights and compute diameter as number of edges only

B. Sum all edge weights and divide by two to get diameter

C. Use BFS twice without any DP or recursion

D. Store and propagate the maximum weighted path length instead of just height counts

Solution

1. Step 1: Understand weighted tree diameter difference

   Diameter must consider edge weights, so heights become max weighted path lengths.

2. Step 2: Modify DP to accumulate weights

   Instead of counting edges, add edge weights when computing subtree heights and update max diameter accordingly.

3. Final Answer:

   Store and propagate the maximum weighted path length instead of just height counts -> Option D

4. Quick Check:

   Weighted diameter needs weighted path sums [OK]

Hint: Accumulate edge weights in DP for weighted diameter [OK]

Common Mistakes:

• Ignoring weights leads to wrong diameter
• Using BFS without weights misses weighted paths
• Incorrect formula dividing total weights