Maximum Path Sum in Binary Tree in DSA Rust - Time & Space Complexity

Time Complexity: Maximum Path Sum in Binary Tree

O(n)

Understanding Time Complexity

We want to understand how the time needed to find the maximum path sum in a binary tree changes as the tree grows.

How does the number of steps grow when the tree has more nodes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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(left + right + n.val);
            left.max(right) + n.val
        } else {
            0
        }
    }

    let mut max_sum = i32::MIN;
    helper(root, &mut max_sum);
    max_sum
}
    

This code finds the maximum sum of any path in a binary tree by visiting each node once and calculating sums from children.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls visiting each node once.
• How many times: Once per node in the tree.

How Execution Grows With Input

Each node is visited once, so the total steps grow directly with the number of nodes.

Input Size (n)	Approx. Operations
10	About 10 visits
100	About 100 visits
1000	About 1000 visits

Pattern observation: The work grows in a straight line with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the maximum path sum grows directly with the number of nodes in the tree.

Common Mistake

[X] Wrong: "The recursion visits some nodes multiple times, so it must be slower than O(n)."

[OK] Correct: Each node is visited exactly once, and results are combined on the way back, so no repeated work happens.

Interview Connect

Understanding this time complexity shows you can analyze recursive tree problems efficiently, a key skill in many coding challenges.

Self-Check

"What if we changed the tree to a linked list (all nodes have only one child)? How would the time complexity change?"

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