DP on Trees Maximum Path Sum in DSA Rust - Build from Scratch

DP on Trees Maximum Path Sum

📖 Scenario: You are working with a tree structure representing a network of connected nodes. Each node has a value that can be positive or negative. You want to find the maximum sum of values along any path in this tree. A path can start and end at any nodes but must follow the tree connections.

🎯 Goal: Build a Rust program that uses dynamic programming on trees to find the maximum path sum. You will create the tree, set up helper variables, write the core logic to compute the maximum path sum, and finally print the result.

📋 What You'll Learn

Create a tree using adjacency lists with given edges and node values

Use a helper variable to track the global maximum path sum

Implement a recursive function to compute maximum path sums using DP on trees

Print the maximum path sum found in the tree

💡 Why This Matters

🌍 Real World

Finding maximum path sums in trees is useful in network analysis, biology (e.g., phylogenetic trees), and optimizing routes in hierarchical data.

💼 Career

Understanding dynamic programming on trees is important for software engineers working on complex data structures, algorithms, and performance optimization.

1

Create the tree data structure

Create a vector called values with these exact integers: [1, -2, 3, 4, -5, 6]. Then create an adjacency list called tree as a vector of vectors of usize with 6 empty vectors. Add edges to tree to represent these connections: between nodes 0 and 1, nodes 0 and 2, nodes 1 and 3, nodes 1 and 4, and nodes 2 and 5.

DSA Rust

let values = vec![1, -2, 3, 4, -5, 6];
let mut tree: Vec> = vec![vec![]; 6];
// Add edges to tree below
// Your code here

Hint

Use a vector of vectors to represent the tree connections. Remember to add edges in both directions since the tree is undirected.

2

Add a helper variable for the maximum path sum

Create a mutable variable called max_sum and set it to i32::MIN to track the global maximum path sum.

DSA Rust

let values = vec![1, -2, 3, 4, -5, 6];
let mut tree: Vec> = vec![vec![]; 6];
// edges added here
let mut max_sum = i32::MIN;
// Your code here

Hint

Use i32::MIN to start with the smallest possible integer.

3

Implement the recursive DP function

Write a recursive function called dfs that takes node: usize, parent: usize, tree: &Vec>, values: &Vec, and max_sum: &mut i32. It returns an i32 representing the maximum path sum starting from node. The function should compute the maximum sums from children, update max_sum with the best path through node, and return the maximum sum path extending from node upwards. Use this function to start DFS from node 0 with parent 6 (an invalid index).

DSA Rust

fn dfs(node: usize, parent: usize, tree: &Vec>, values: &Vec, max_sum: &mut i32) -> i32 {
    // Your code here
}

dfs(0, 6, &tree, &values, &mut max_sum);

Hint

Use DFS to explore children. Keep track of the top two maximum child sums that are positive. Update the global max with the sum of node value plus these two child sums. Return the node value plus the maximum child sum for parent use.

4

Print the maximum path sum

Write a print statement to display the value of max_sum.

DSA Rust

println!("Maximum path sum: {}", max_sum);

Hint

Use println!("Maximum path sum: {}", max_sum); to display the result.

Practice

1. What does the "maximum path sum" in a tree represent in the context of DP on Trees Maximum Path Sum?

easy

A. The largest sum of node values along any path between two nodes in the tree

B. The sum of all node values in the tree

C. The maximum value of a single node in the tree

D. The number of nodes in the longest path of the tree

Solution

1. Step 1: Understand the definition of path sum

   The path sum is the total of node values along a path connecting two nodes in the tree.

2. Step 2: Identify the maximum path sum meaning

   The maximum path sum is the largest such sum among all possible paths in the tree.

3. Final Answer:

   The largest sum of node values along any path between two nodes in the tree -> Option A

4. Quick Check:

   Maximum path sum = largest node value sum on any path [OK]

Hint: Maximum path sum is about sums on paths, not single nodes [OK]

Common Mistakes:

• Confusing maximum path sum with sum of all nodes
• Thinking it is just the max node value
• Counting number of nodes instead of sum

2. Which of the following Rust function signatures is correct for a recursive function that calculates the maximum path sum in a binary tree node defined as struct TreeNode { val: i32, left: Option, right: Option }?

easy

A. fn max_path_sum(node: Box<TreeNode>) -> Option<i32>

B. fn max_path_sum(node: TreeNode) -> i32

C. fn max_path_sum(node: &Option) -> i32

D. fn max_path_sum(node: &TreeNode) -> Option<i32>

Solution

1. Step 1: Understand ownership and borrowing in Rust

   Recursive tree functions usually borrow nodes to avoid ownership transfer and cloning.

2. Step 2: Match function signature to tree node type

   The tree node is wrapped in Option, so borrowing it as &Option is correct for recursion.

3. Final Answer:

   fn max_path_sum(node: &Option) -> i32 -> Option C

4. Quick Check:

   Borrow Option and return i32 [OK]

Hint: Use &Option to borrow nodes recursively [OK]

Common Mistakes:

• Taking ownership of node causing move errors
• Returning Option<i32> unnecessarily
• Using &TreeNode instead of &Option

3. Given the following Rust code snippet for maximum path sum in a binary tree, what is the output printed?

struct TreeNode {
    val: i32,
    left: Option,
    right: Option,
}

fn max_gain(node: &Option, max_sum: &mut i32) -> i32 {
    if let Some(n) = node {
        let left_gain = max_gain(&n.left, max_sum).max(0);
        let right_gain = max_gain(&n.right, max_sum).max(0);
        let price_newpath = n.val + left_gain + right_gain;
        *max_sum = (*max_sum).max(price_newpath);
        n.val + left_gain.max(right_gain)
    } else {
        0
    }
}

fn max_path_sum(root: &Option) -> i32 {
    let mut max_sum = i32::MIN;
    max_gain(root, &mut max_sum);
    max_sum
}

fn main() {
    let root = Some(Box::new(TreeNode {
        val: -10,
        left: Some(Box::new(TreeNode {
            val: 9,
            left: None,
            right: None,
        })),
        right: Some(Box::new(TreeNode {
            val: 20,
            left: Some(Box::new(TreeNode {
                val: 15,
                left: None,
                right: None,
            })),
            right: Some(Box::new(TreeNode {
                val: 7,
                left: None,
                right: None,
            })),
        })),
    }));
    println!("{}", max_path_sum(&root));
}

medium

A. 42

B. 35

C. 27

D. 20

Solution

1. Step 1: Calculate max path sums for each subtree

   Calculate max gains: - Left subtree (9): max gain = 9 - Right subtree: - Left child (15): max gain = 15 - Right child (7): max gain = 7 - Node 20: max gain = 20 + max(15,7) = 35 - Max path through 20: 20 + 15 + 7 = 42

2. Step 2: Calculate max path including root (-10)

   Root max gain = -10 + max(9, 35) = 25 Max path sum is max of: - Left subtree max: 9 - Right subtree max: 42 - Root path: 25 So max path sum = 42

3. Final Answer:

   42 -> Option A

4. Quick Check:

   Max path sum = 42 [OK]

Hint: Add node value plus max left/right gains, track max path [OK]

Common Mistakes:

• Ignoring negative values and not using max(0, gain)
• Adding gains incorrectly for path through root
• Confusing max gain with max path sum

4. The following Rust function attempts to compute the maximum path sum in a binary tree but contains a bug. What is the bug?

fn max_gain(node: &Option, max_sum: &mut i32) -> i32 {
    if let Some(n) = node {
        let left_gain = max_gain(&n.left, max_sum);
        let right_gain = max_gain(&n.right, max_sum);
        let price_newpath = n.val + left_gain + right_gain;
        *max_sum = (*max_sum).max(price_newpath);
        n.val + left_gain.max(right_gain)
    } else {
        0
    }
}

medium

A. It incorrectly updates max_sum inside recursion

B. It does not handle empty tree nodes

C. It returns the sum of left and right gains instead of max

D. It does not ignore negative gains by using max(0, gain)

Solution

1. Step 1: Analyze gain calculation for left and right subtrees

   The function uses left_gain and right_gain directly without ignoring negative values, which can reduce the path sum.

2. Step 2: Identify the fix for negative gains

   Correct approach is to use max_gain = max(0, recursive_gain) to ignore negative paths that reduce sum.

3. Final Answer:

   It does not ignore negative gains by using max(0, gain) -> Option D

4. Quick Check:

   Use max(0, gain) to avoid negative path sums [OK]

Hint: Always ignore negative gains with max(0, gain) [OK]

Common Mistakes:

• Forgetting to ignore negative subtree sums
• Updating max_sum incorrectly
• Returning wrong values from recursion

5. You want to modify the maximum path sum algorithm to find the maximum path sum that must include the root node in a binary tree. Which change should you make to the existing DP on Trees Maximum Path Sum approach?

hard

A. Use a global variable to store max sum and update it only when the path includes root and both children

B. Return the sum of the root value plus the maximum gain from either left or right subtree only, and track max sum accordingly

C. Calculate max path sum as usual but ignore paths that do not include the root node

D. Traverse the tree twice: once for max path sum and once for root-included path sum, then compare

Solution

1. Step 1: Understand the root-included path sum requirement

   The path must include the root node, so the path is root plus one subtree path (left or right), not both sides.

2. Step 2: Modify the gain calculation accordingly

   Return root.val + max(left_gain, right_gain) ignoring negative gains, and track max sum as this value.

3. Final Answer:

   Return the sum of the root value plus the maximum gain from either left or right subtree only, and track max sum accordingly -> Option B

4. Quick Check:

   Root-included max path = root + max(left, right) [OK]

Hint: Include root plus max one subtree gain only [OK]

Common Mistakes:

• Adding both left and right gains when root must be included
• Ignoring negative gains
• Trying to filter paths after calculation