DP on Trees Maximum Path Sum in DSA Rust - Time & Space Complexity

Time Complexity: DP on Trees Maximum Path Sum

O(n)

Understanding Time Complexity

We want to understand how the time needed grows when finding the maximum path sum in a tree using dynamic programming.

How does the number of steps change as the tree gets bigger?

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 values along any path in a binary tree using recursion and dynamic programming.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls visiting each node once.
• How many times: Each node is visited exactly one time.

How Execution Grows With Input

As the tree grows, the number of nodes increases, and the function visits each node once.

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

Pattern observation: The steps grow directly with the number of nodes, so doubling nodes doubles work.

Final Time Complexity

Time Complexity: O(n)

This means the time needed grows linearly with the number of nodes in the tree.

Common Mistake

[X] Wrong: "Because the function calls itself twice, the time is O(2^n)."

[OK] Correct: Each node is visited once, not repeatedly, so calls do not multiply exponentially.

Interview Connect

Understanding this helps you explain how recursion and dynamic programming work together efficiently on trees.

Self-Check

"What if the tree was very unbalanced, like a linked list? How would the time complexity change?"

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