DP on Trees Maximum Path Sum in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to define the TreeNode struct with optional children.

DSA Rust

struct TreeNode {
    val: i32,
    left: Option<Box<TreeNode>>, 
    right: Option<Box<TreeNode>>, 
}

impl TreeNode {
    fn new(val: i32) -> Self {
        TreeNode {
            val,
            left: [1],
            right: None,
        }
    }
}

Drag options to blanks, or click blank then click option'

ANone

BSome(Box::new(TreeNode::new(val)))

CSome(TreeNode::new(val))

DSome(Box::new(val))

Attempts:

3 left

💡 Hint

Common Mistakes

Using Some without Box wrapping causes type errors.

Assigning a value directly instead of Option.

The left child is initially empty, so we use None to indicate no child.

2fill in blank

medium

Complete the code to calculate the maximum path sum including the current node and one child.

DSA Rust

fn max_gain(node: &Option<Box<TreeNode>>, 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);
        return n.val + [1].max([2]);
    }
    0
}

Drag options to blanks, or click blank then click option'

Aright_gain

Bleft_gain

Cleft_gain + right_gain

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Adding both left and right gains when returning causes incorrect path sums.

Returning zero always ignores node values.

We return the node's value plus the maximum gain from either left or right child to continue the path upwards.

3fill in blank

hard

Fix the error in the recursive call to handle the right child correctly.

DSA Rust

fn max_gain(node: &Option<Box<TreeNode>>, 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(&[1], max_sum).max(0);
        let price_newpath = n.val + left_gain + right_gain;
        *max_sum = (*max_sum).max(price_newpath);
        return n.val + left_gain.max(right_gain);
    }
    0
}

Drag options to blanks, or click blank then click option'

An.right

Bn.left

Cnode.right

Dnode.left

Attempts:

3 left

💡 Hint

Common Mistakes

Using node.right causes errors because node is an Option.

Using n.left instead of right child.

The right child is accessed via n.right inside the Some(n) block.

4fill in blank

hard

Fill both blanks to complete the main function that initializes max_sum and calls max_gain.

DSA Rust

fn max_path_sum(root: Option<Box<TreeNode>>) -> i32 {
    let mut max_sum = [1];
    max_gain(&root, &mut max_sum);
    [2]
}

Drag options to blanks, or click blank then click option'

Ai32::MIN

Bmax_sum

C0

Droot.val

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing max_sum with 0 misses negative path sums.

Returning root.val instead of max_sum.

Initialize max_sum with the smallest integer to handle negative values, then return max_sum after computation.

5fill in blank

hard

Fill all three blanks to complete the recursive function that updates max_sum and returns the max gain.

DSA Rust

fn max_gain(node: &Option<Box<TreeNode>>, max_sum: &mut i32) -> i32 {
    if let Some(n) = node {
        let left_gain = max_gain(&n.left, max_sum).max([1]);
        let right_gain = max_gain(&n.right, max_sum).max([2]);
        let price_newpath = n.val + left_gain + right_gain;
        *max_sum = (*max_sum).max([3]);
        return n.val + left_gain.max(right_gain);
    }
    0
}

Drag options to blanks, or click blank then click option'

A0

Bn.val

Cprice_newpath

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 instead of 0 causes incorrect gain calculations.

Updating max_sum with node.val only misses full path sums.

We ignore negative gains by comparing with 0, and update max_sum with the new path price.

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