Maximum Path Sum in Binary Tree in DSA Rust - Build from Scratch

Maximum Path Sum in Binary Tree

📖 Scenario: Imagine you are working with a tree of numbers representing different paths in a park. Each node is a checkpoint with a score. You want to find the path through the park that gives the highest total score. This path can start and end at any checkpoint but must follow the tree connections.

🎯 Goal: Build a Rust program that finds the maximum path sum in a binary tree. The program will create the tree, set up a helper variable, implement the logic to find the maximum path sum, and print the result.

📋 What You'll Learn

Create a binary tree with the exact structure and values given

Add a mutable variable to track the maximum path sum

Implement a recursive function to calculate the maximum path sum

Print the maximum path sum after processing the tree

💡 Why This Matters

🌍 Real World

Finding the maximum path sum in a tree is useful in network routing, game development for scoring paths, and analyzing hierarchical data.

💼 Career

Understanding tree traversal and recursion is essential for software engineering roles involving data structures, algorithms, and problem-solving.

1

Create the Binary Tree Structure

Define a TreeNode struct with val, left, and right fields. Then create a binary tree with root value 1, left child 2, and right child 3 using Box and Option.

DSA Rust

// Define TreeNode struct and create the tree
// Your code here

Hint

Use struct to define the tree node. Use Option> for left and right children. Create the root node with value 1, and add left and right children with values 2 and 3.

2

Add a Mutable Variable to Track Maximum Path Sum

Inside main, create a mutable variable called max_sum and set it to i32::MIN to track the maximum path sum found.

DSA Rust

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

fn main() {
    let root = Some(Box::new(TreeNode {
        val: 1,
        left: Some(Box::new(TreeNode {
            val: 2,
            left: None,
            right: None,
        })),
        right: Some(Box::new(TreeNode {
            val: 3,
            left: None,
            right: None,
        })),
    }));

    // Create a mutable variable max_sum and set it to i32::MIN
    // Your code here
}

Hint

Use let mut max_sum = i32::MIN; inside main to track the highest path sum.

3

Implement Recursive Function to Find Maximum Path Sum

Write a recursive function called max_gain that takes a reference to an Option> and a mutable reference to max_sum. It returns the maximum gain from the current node. Update max_sum with the maximum path sum found including the current node.

DSA Rust

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

fn main() {
    let root = Some(Box::new(TreeNode {
        val: 1,
        left: Some(Box::new(TreeNode {
            val: 2,
            left: None,
            right: None,
        })),
        right: Some(Box::new(TreeNode {
            val: 3,
            left: None,
            right: None,
        })),
    }));

    let mut max_sum = i32::MIN;

    // Implement max_gain function here
    // Your code here
}

Hint

Use recursion to get max gain from left and right children. Ignore negative gains by using .max(0). Update max_sum if current path sum is higher. Return node value plus max gain from one child.

4

Print the Maximum Path Sum

Add a println! statement to print the value of max_sum after calling max_gain.

DSA Rust

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;

        if price_newpath > *max_sum {
            *max_sum = price_newpath;
        }

        n.val + left_gain.max(right_gain)
    } else {
        0
    }
}

fn main() {
    let root = Some(Box::new(TreeNode {
        val: 1,
        left: Some(Box::new(TreeNode {
            val: 2,
            left: None,
            right: None,
        })),
        right: Some(Box::new(TreeNode {
            val: 3,
            left: None,
            right: None,
        })),
    }));

    let mut max_sum = i32::MIN;

    max_gain(&root, &mut max_sum);

    // Print the max_sum
    // Your code here
}

Hint

Use println!("{}", max_sum); to display the maximum path sum after the recursive calls.

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