Maximum Path Sum in Binary Tree in DSA Go - 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 grows as the tree gets bigger.

Specifically, how does the number of steps change when the number of nodes increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func maxPathSum(root *TreeNode) int {
    maxSum := math.MinInt32
    var dfs func(node *TreeNode) int
    dfs = func(node *TreeNode) int {
        if node == nil {
            return 0
        }
        left := max(0, dfs(node.Left))
        right := max(0, dfs(node.Right))
        maxSum = max(maxSum, left + right + node.Val)
        return max(left, right) + node.Val
    }
    dfs(root)
    return maxSum
}

func max(a, b int) int {
    if a > b {
        return a
    }
    return b
}
    

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

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive depth-first search (dfs) visits each node once.
• How many times: Exactly 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 linearly as the tree size increases.

Final Time Complexity

Time Complexity: O(n)

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

Common Mistake

[X] Wrong: "The recursion visits nodes multiple times, so the time is more than linear."

[OK] Correct: Each node is visited once, and results from children are reused immediately, so no repeated work happens.

Interview Connect

Understanding this linear time complexity helps you explain why your solution is efficient and suitable for large trees, a skill valued in interviews.

Self-Check

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

Practice

1. What does the term Maximum Path Sum in a binary tree mean?

easy

A. The largest sum of values along any path between two nodes in the tree

B. The sum of all node values in the tree

C. The sum of values from root to the leftmost leaf

D. The smallest sum of values along any path in the tree

Solution

1. Step 1: Understand the path definition

   The path can start and end at any two nodes, not necessarily root or leaf.

2. Step 2: Identify maximum sum meaning

   We look for the path with the highest total sum of node values.

3. Final Answer:

   The largest sum of values along any path between two nodes in the tree -> Option A

4. Quick Check:

   Maximum path sum means highest sum between any two nodes [OK]

Hint: Maximum path sum is any path with highest total node values [OK]

Common Mistakes:

• Thinking path must start at root
• Confusing maximum path sum with total tree sum
• Assuming path must end at leaf

2. Which of the following Go function signatures is correct for a function that calculates maximum path sum in a binary tree node defined as type TreeNode struct { Val int; Left, Right *TreeNode }?

easy

A. func maxPathSum(root *TreeNode) int

B. func maxPathSum(root TreeNode) int

C. func maxPathSum(root *TreeNode) *int

D. func maxPathSum(root *TreeNode) string

Solution

1. Step 1: Check parameter type

   Function should accept pointer to TreeNode to avoid copying the whole tree.

2. Step 2: Check return type

   Maximum path sum is an integer value, so return type should be int.

3. Final Answer:

   func maxPathSum(root *TreeNode) int -> Option A

4. Quick Check:

   Pointer input and int output for max path sum [OK]

Hint: Use pointer receiver and int return for tree max path sum [OK]

Common Mistakes:

• Passing TreeNode by value causing copies
• Returning pointer instead of int
• Returning string instead of int

3. Given the following Go code snippet, what is the printed output?

type TreeNode struct {
    Val   int
    Left  *TreeNode
    Right *TreeNode
}

func maxPathSum(root *TreeNode) int {
    maxSum := -1 << 31
    var maxGain func(node *TreeNode) int
    maxGain = func(node *TreeNode) int {
        if node == nil {
            return 0
        }
        leftGain := max(0, maxGain(node.Left))
        rightGain := max(0, maxGain(node.Right))
        priceNewPath := node.Val + leftGain + rightGain
        if priceNewPath > maxSum {
            maxSum = priceNewPath
        }
        return node.Val + max(leftGain, rightGain)
    }
    maxGain(root)
    return maxSum
}

func max(a, b int) int {
    if a > b {
        return a
    }
    return b
}

func main() {
    root := &TreeNode{Val: -10}
    root.Left = &TreeNode{Val: 9}
    root.Right = &TreeNode{Val: 20}
    root.Right.Left = &TreeNode{Val: 15}
    root.Right.Right = &TreeNode{Val: 7}
    fmt.Println(maxPathSum(root))
}

medium

A. 27

B. 42

C. 35

D. 20

Solution

1. Step 1: Trace maxGain for each node

   Calculate max gain from left and right subtree ignoring negatives: - Node 9: maxGain = 9 - Node 15: maxGain = 15 - Node 7: maxGain = 7 - Node 20: maxGain = 20 + max(15,7) = 35 - Root -10: maxGain = -10 + max(9,35) = 25 Check maxSum updates: - At node 9: maxSum = 9 - At node 15: maxSum = 15 - At node 7: maxSum = 15 (unchanged) - At node 20: priceNewPath = 20 + 15 + 7 = 42, maxSum = 42 - At root: priceNewPath = -10 + 9 + 35 = 34 (maxSum stays 42)

2. Step 2: Return maxSum

   maxSum is 42 after full traversal.

3. Final Answer:

   42 -> Option B

4. Quick Check:

   Max path sum = 42 [OK]

Hint: Add node value plus max gains from left and right, track max sum [OK]

Common Mistakes:

• Not ignoring negative gains
• Returning max gain instead of max path sum
• Miscomputing path sums including negative nodes

4. The following Go function attempts to compute the maximum path sum in a binary tree but has a bug. What is the bug?

func maxPathSum(root *TreeNode) int {
    maxSum := -1 << 31
    var maxGain func(node *TreeNode) int
    maxGain = func(node *TreeNode) int {
        if node == nil {
            return 0
        }
        leftGain := maxGain(node.Left)
        rightGain := maxGain(node.Right)
        priceNewPath := node.Val + leftGain + rightGain
        if priceNewPath > maxSum {
            maxSum = priceNewPath
        }
        return node.Val + max(leftGain, rightGain)
    }
    maxGain(root)
    return maxSum
}

medium

A. It does not handle nil root

B. It returns maxSum before traversal

C. It does not ignore negative gains from left or right subtrees

D. It uses incorrect initial value for maxSum

Solution

1. Step 1: Analyze gain calculation

   The function uses leftGain and rightGain directly without ignoring negative values, which can reduce the path sum.

2. Step 2: Identify fix

   To fix, use max(0, leftGain) and max(0, rightGain) to ignore negative gains.

3. Final Answer:

   It does not ignore negative gains from left or right subtrees -> Option C

4. Quick Check:

   Negative gains must be ignored for correct max path sum [OK]

Hint: Always ignore negative subtree sums by max(0, gain) [OK]

Common Mistakes:

• Not ignoring negative subtree sums
• Wrong initial maxSum value
• Returning before traversal completes

5. You want to find the maximum path sum in a binary tree where some nodes have negative values. Which approach correctly handles this scenario? A) Use a recursive function that returns the maximum gain from each node, ignoring negative gains by taking max(0, gain), and update a global max sum including both left and right gains plus node value. B) Use a recursive function that sums all node values and returns the total. C) Use a breadth-first search to find the maximum node value and return it. D) Use a recursive function that only considers paths from root to leaf nodes.

hard

A. Consider only root-to-leaf paths recursively

B. Sum all node values recursively and return total

C. Find maximum node value using BFS and return it

D. Use recursion with max(0, gain) and update max sum including both children

Solution

1. Step 1: Understand negative values impact

   Negative node values can reduce path sums, so ignoring negative gains is necessary.

2. Step 2: Choose correct approach

   Use recursion with max(0, gain) and update max sum including both children uses recursion with max(0, gain) to ignore negative paths and updates max sum considering both left and right children plus current node, which correctly finds maximum path sum anywhere in the tree.

3. Final Answer:

   Use recursion with max(0, gain) and update max sum including both children -> Option D

4. Quick Check:

   Ignore negative gains and consider all paths for max sum [OK]

Hint: Ignore negative gains and consider all paths recursively [OK]

Common Mistakes:

• Summing all nodes ignoring path constraints
• Only considering root-to-leaf paths
• Returning max node value instead of path sum