Maximum Path Sum in Binary Tree in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to define the TreeNode struct with optional left and right children.

DSA Rust

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

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))

DBox::new(None)

Attempts:

3 left

💡 Hint

Common Mistakes

Using Some without Box causes type errors.

Initializing children with values instead of None.

The left child should be initialized as None when creating a new node without children.

2fill in blank

medium

Complete the code to update the maximum path sum during recursion.

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).[1](price_newpath);
        n.val + left_gain.max(right_gain)
    } else {
        0
    }
}

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Amin

Bmax

Cabs

Dpow

Attempts:

3 left

💡 Hint

Common Mistakes

Using min instead of max causes wrong results.

Forgetting to update max_sum.

We update max_sum to the maximum of current max_sum and the new path sum.

3fill in blank

hard

Fix the error in the recursive function to correctly handle the base case.

DSA Rust

fn max_gain(node: &Option<Box<TreeNode>>, max_sum: &mut i32) -> i32 {
    if node.is_none() {
        return [1];
    }
    let n = node.as_ref().unwrap();
    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)
}

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A0

B-1

Cn.val

Di32::MIN

Attempts:

3 left

💡 Hint

Common Mistakes

Returning n.val causes panic on None.

Returning negative values reduces max path sum incorrectly.

The base case returns 0 because a null node contributes no gain.

4fill in blank

hard

Fill both blanks to create a function that initializes max_sum and calls the recursive helper.

DSA Rust

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

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Ai32::MIN

B0

Cmax_sum

Droot.val

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing max_sum to 0 causes errors with negative values.

Not returning max_sum causes function to return unit type.

max_sum starts at the smallest integer to allow any path sum to update it. The function returns max_sum after recursion.

5fill in blank

hard

Fill all three blanks to complete the recursive max_gain function with correct logic.

DSA Rust

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

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

Amax

Bmin

Dpow

Attempts:

3 left

💡 Hint

Common Mistakes

Using min instead of max causes wrong path sums.

Not ignoring negative gains leads to smaller sums.

We use max(0) to ignore negative gains, and max to choose the larger gain to continue the path.

Practice

1. What does the "Maximum Path Sum in Binary Tree" problem ask you to find?

easy

A. The longest path length in the binary tree

B. The number of leaf nodes in the binary tree

C. The highest sum of values along any path in the binary tree

D. The minimum value stored in the binary tree

Solution

1. Step 1: Understand the problem statement

   The problem asks for the maximum sum of node values along any path in the binary tree. A path can start and end at any node, but must be connected through parent-child links.

2. Step 2: Clarify what a path means

   A path can go downwards or upwards but cannot revisit nodes. The goal is to find the path with the largest sum of node values.

3. Final Answer:

   The highest sum of values along any path in the binary tree -> Option C

4. Quick Check:

   Maximum path sum means largest sum of connected nodes [OK]

Hint: Focus on sums of connected nodes, not just length [OK]

Common Mistakes:

• Confusing path length with path sum
• Thinking path must start at root
• Ignoring negative values in nodes

2. Which Rust function signature correctly represents a function to find 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(root: Option) -> i32

B. fn max_path_sum(root: TreeNode) -> Option<i32>

C. fn max_path_sum(root: &TreeNode) -> i32

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

Solution

1. Step 1: Understand Rust ownership and option types

   The tree root is usually wrapped in Option to allow for empty nodes and heap allocation.

2. Step 2: Choose correct function signature

   The function should accept Option and return i32 for the maximum path sum. Passing by value is common for ownership transfer.

3. Final Answer:

   fn max_path_sum(root: Option) -> i32 -> Option A

4. Quick Check:

   Option input and i32 output [OK]

Hint: Use Option for nullable tree nodes [OK]

Common Mistakes:

• Using TreeNode without Box causes size errors
• Returning Option<i32> unnecessarily
• Passing root by reference when ownership is needed

3. Given the following Rust code snippet for maximum path sum, what will be the output?

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

fn max_path_sum(root: Option) -> i32 {
    fn helper(node: &Option, max_sum: &mut i32) -> i32 {
        if let Some(n) = node {
            let left = helper(&n.left, max_sum).max(0);
            let right = helper(&n.right, max_sum).max(0);
            *max_sum = (*max_sum).max(n.val + left + right);
            n.val + left.max(right)
        } else {
            0
        }
    }
    let mut max_sum = i32::MIN;
    helper(&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. 20

B. 35

C. 27

D. 42

Solution

1. Step 1: Trace recursive helper calls and calculate sums

   Calculate max path sums for left and right subtrees, ignoring negative sums by max(0). For node 20, left=15, right=7, sum=20+15+7=42. For root -10, left max=9, right max=42, total= -10+9+42=41 but max_sum is 42 from node 20 subtree.

2. Step 2: Determine maximum path sum found

   The maximum path sum is 42 from path 15 -> 20 -> 7.

3. Final Answer:

   42 -> Option D

4. Quick Check:

   Max path sum = 42 [OK]

Hint: Ignore negative sums by max(0) to maximize path sum [OK]

Common Mistakes:

• Including negative sums in path
• Confusing max path sum with max root-to-leaf sum
• Miscomputing sums for left and right subtrees

4. Identify the bug in this Rust function snippet for maximum path sum and select the fix:

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

medium

A. Add .max(0) to left and right to ignore negative sums

B. Change return type to Option<i32>

C. Remove max_sum parameter and return max sum directly

D. Replace n.val + left.max(right) with n.val + left + right

Solution

1. Step 1: Identify problem with negative sums

   Without .max(0), negative sums from left or right subtrees reduce the path sum, which is incorrect for maximum path sum calculation.

2. Step 2: Fix by ignoring negative sums

   Adding .max(0) to left and right ensures negative sums are treated as zero, improving the maximum path sum calculation.

3. Final Answer:

   Add .max(0) to left and right to ignore negative sums -> Option A

4. Quick Check:

   Ignore negative sums with max(0) [OK]

Hint: Always ignore negative subtree sums with max(0) [OK]

Common Mistakes:

• Not handling negative values correctly
• Changing return type unnecessarily
• Summing left and right twice

5. You want to modify the maximum path sum function to also return the path nodes (values) that produce the maximum sum. Which approach is best to implement this in Rust?

hard

A. Return only max_sum and reconstruct path by traversing tree again after computation

B. Use a helper function returning a tuple of (max_sum, path_vec) and update global max with path

C. Store paths in a global mutable vector and update it inside recursion without returning

D. Use a stack to store nodes during recursion and pop when max sum is found

Solution

1. Step 1: Understand need to track path along with sums

   To return the path nodes, the helper must return both the max sum and the path that produces it.

2. Step 2: Use tuple return and update global max with path

   Returning (max_sum, path_vec) allows updating the global max sum and storing the corresponding path during recursion.

3. Final Answer:

   Use a helper function returning a tuple of (max_sum, path_vec) and update global max with path -> Option B

4. Quick Check:

   Return sum and path together for tracking [OK]

Hint: Return both sum and path as tuple from recursion [OK]

Common Mistakes:

• Using global mutable state without return
• Recomputing path after sum calculation
• Using stack incorrectly for path tracking