Maximum Path Sum in Binary Tree in DSA Rust - Execution Trace

Concept Flow - Maximum Path Sum in Binary Tree

Start at root node

↓

Calculate max path sum from left subtree

↓

Calculate max path sum from right subtree

↓

Calculate max path sum passing through current node

↓

Update global max if current path sum is greater

↓

Return max path sum including current node and one subtree

↓

Repeat for all nodes

↓

Final max path sum found

Start at root, recursively find max path sums from left and right subtrees, update global max with paths through current node, return max path sum including current node and one subtree, repeat for all nodes.

Execution Sample

DSA Rust

fn max_path_sum(node: Option<&Box<TreeNode>>, max_sum: &mut i32) -> i32 {
    if node.is_none() { return 0; }
    let n = node.unwrap();
    let left = max_path_sum(n.left.as_ref(), max_sum).max(0);
    let right = max_path_sum(n.right.as_ref(), max_sum).max(0);
    *max_sum = (*max_sum).max(n.val + left + right);
    n.val + left.max(right)
}

Recursively computes max path sum in binary tree, updating global max sum.

Execution Table

Step	Operation	Node Visited	Left Max	Right Max	Current Path Sum	Global Max	Return Value	Visual State
1	Visit node	1	0	0	1	1	1	Node 1 (val=1)
2	Visit node	2	0	0	2	2	2	Node 2 (val=2)
3	Visit node	4	0	0	4	4	4	Node 4 (val=4)
4	Return to node	2	4	0	6	6	6	Node 2 with left=4
5	Visit node	5	0	0	5	6	5	Node 5 (val=5)
6	Return to node	2	4	5	11	11	9	Node 2 with left=4, right=5
7	Return to node	1	9	0	10	11	10	Node 1 with left=9
8	Visit node	3	0	0	-3	11	0	Node 3 (val=-3)
9	Return to node	1	9	0	10	11	10	Node 1 with left=9, right=0
10	End	-	-	-	-	11	-	Final max path sum = 11

💡 All nodes visited, recursion ends with global max sum = 11

Variable Tracker

Variable	Start	After Step 3	After Step 6	After Step 9	Final
max_sum	-∞	4	11	11	11
left	0	4	4	9	9
right	0	0	5	0	0
return_value	-	4	9	10	10

Key Moments - 3 Insights

Why do we use max(0, left) and max(0, right) when calculating left and right max?

Because negative path sums reduce the total, we ignore them by taking max with 0. See execution_table rows 3 and 8 where negative or zero values are replaced by 0.

Why do we update the global max with n.val + left + right instead of just one side?

Because the path can pass through the current node connecting left and right subtrees. This is shown in execution_table row 6 where global max is updated with sum including both sides.

What does the return value represent at each node?

It represents the max path sum including the current node and one subtree (left or right), to be used by the parent node. See execution_table rows 4, 6, 7, and 9.

Visual Quiz - 3 Questions

Look at the execution_table at step 6, what is the global max after visiting node 2?

A 6

B 11

C 9

D 5

💡 Hint

Check the 'Global Max' column at step 6 in execution_table

At which step does the left max value for node 1 become 9?

A Step 7

B Step 6

C Step 4

D Step 9

💡 Hint

Look at the 'Left Max' column for node 1 in execution_table

If node 3 had a positive value instead of -3, how would the global max change at step 9?

A It would decrease

B It would stay the same

C It would increase

D It would become zero

💡 Hint

Positive right max would add to current path sum, increasing global max (see variable_tracker for right values)

Concept Snapshot

Maximum Path Sum in Binary Tree:
- Recursively find max path sums from left and right children
- Ignore negative sums by max(0, child_sum)
- Update global max with node.val + left + right
- Return node.val + max(left, right) to parent
- Final global max is the answer

Full Transcript

This concept finds the maximum path sum in a binary tree by visiting each node recursively. At each node, it calculates the maximum path sum from the left and right subtrees, ignoring negative sums by taking max with zero. It then updates a global maximum sum with the sum of the current node's value plus the left and right max sums, representing a path passing through the node. The function returns the maximum sum including the current node and one subtree to its parent. This process repeats for all nodes, and the global maximum sum at the end is the maximum path sum in the tree.

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