DP on Trees Maximum Path Sum in DSA Go - 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.

// Tree node structure
 type TreeNode struct {
     Val   int
     Left  *TreeNode
     Right *TreeNode
 }

func maxPathSum(root *TreeNode) int {
    maxSum := root.Val
    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, node.Val + left + right)
        return node.Val + max(left, right)
    }
    dfs(root)
    return maxSum
}

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

This code finds the maximum sum of values along any path in a binary tree by visiting each node once.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive depth-first search (dfs) visiting each node once.
• How many times: Each node is visited exactly one time during the recursion.

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 number of operations grows roughly in direct proportion to the number of nodes.

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 only once, not repeatedly, so the total work is proportional to the number of nodes, not exponential.

Interview Connect

Understanding this linear time complexity helps you explain how efficient tree traversals with dynamic programming work, a common skill interviewers look for.

Self-Check

"What if we changed the tree to a graph with cycles? How would the time complexity change?"

Practice

1. What does the "Maximum Path Sum" in a tree represent in the context of dynamic programming on trees?

easy

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

B. The number of nodes in the longest path of the tree

C. The sum of all node values in the tree

D. The maximum value of a single node in the tree

Solution

1. Step 1: Understand the problem definition

   The maximum path sum is about finding the path between any two nodes that gives the highest sum of node values.

2. Step 2: Differentiate from other sums

   It is not just the sum of all nodes or the longest path length, but the maximum sum along any path.

3. Final Answer:

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

4. Quick Check:

   Maximum path sum = largest sum on any path [OK]

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

Common Mistakes:

• Confusing maximum path sum with longest path length
• Thinking it is sum of all nodes
• Considering only single node values

2. Which of the following Go function signatures is correct for a recursive function that returns the maximum path sum starting from a given tree node?

easy

A. func maxPathSum(node int) int

B. func maxPathSum(node TreeNode) string

C. func maxPathSum(node *TreeNode) (int, error)

D. func maxPathSum(node *TreeNode) int

Solution

1. Step 1: Identify correct parameter type

   The function should accept a pointer to TreeNode to handle nil nodes and avoid copying.

2. Step 2: Identify correct return type

   The function returns an integer representing the maximum path sum, so return type int is correct.

3. Final Answer:

   func maxPathSum(node *TreeNode) int -> Option D

4. Quick Check:

   Pointer input and int output for max sum [OK]

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

Common Mistakes:

• Passing node by value instead of pointer
• Returning string or error instead of int
• Using wrong parameter type like int

3. Given the following Go code snippet for a binary tree node and a function to compute maximum path sum, what is the output printed?

package main

import "fmt"

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

func maxPathSumUtil(node *TreeNode, maxSum *int) int {
    if node == nil {
        return 0
    }
    left := max(0, maxPathSumUtil(node.Left, maxSum))
    right := max(0, maxPathSumUtil(node.Right, maxSum))
    *maxSum = max(*maxSum, left + right + node.Val)
    return max(left, right) + node.Val
}

func maxPathSum(root *TreeNode) int {
    maxSum := -1000000
    maxPathSumUtil(root, &maxSum)
    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. 35

C. 42

D. 20

Solution

1. Step 1: Calculate max path sums for subtrees

   Left subtree: 9 (updates maxSum to 9). Right subtree: 15 (maxSum=15), 7 (no change), at 20: left=15, right=7, *maxSum = max(15, 15+7+20)=42, returns max(15,7)+20=35.

2. Step 2: Compare and update maxSum

   At root (-10): left=max(0,9)=9, right=max(0,35)=35, *maxSum = max(42, 9+35+(-10))=max(42,34)=42.

3. Final Answer:

   42 -> Option C

4. Quick Check:

   Max path sum = 42 [OK]

Hint: Add max(0, child sums) and update global max [OK]

Common Mistakes:

• Not ignoring negative sums by using max(0, ...)
• Adding node values incorrectly
• Confusing max path sum with max root-to-leaf sum

4. The following Go function is intended to compute the maximum path sum in a binary tree. What is the bug causing incorrect results?

func maxPathSumUtil(node *TreeNode, maxSum *int) int {
    if node == nil {
        return 0
    }
    left := maxPathSumUtil(node.Left, maxSum)
    right := maxPathSumUtil(node.Right, maxSum)
    *maxSum = max(*maxSum, left + right + node.Val)
    return max(left, right) + node.Val
}

medium

A. It does not ignore negative sums from children before adding

B. It returns the sum of both children instead of max child sum

C. It does not update maxSum correctly

D. It does not handle nil nodes

Solution

1. Step 1: Check handling of negative child sums

   The function uses left and right sums directly without ignoring negative values, which can reduce the path sum.

2. Step 2: Identify fix by ignoring negatives

   To fix, use max(0, left) and max(0, right) to ignore negative sums before adding to node.Val.

3. Final Answer:

   It does not ignore negative sums from children before adding -> Option A

4. Quick Check:

   Ignore negative child sums for correct max path [OK]

Hint: Always ignore negative child sums with max(0, child) [OK]

Common Mistakes:

• Not using max(0, child) to ignore negatives
• Returning sum of both children instead of max child
• Not updating maxSum properly

5. You are given a binary tree where some nodes have negative values. How would you modify the standard DP on trees maximum path sum algorithm to correctly handle this case and find the maximum path sum?

hard

A. Ignore all negative nodes completely from the tree

B. Use max(0, child sums) to ignore negative paths and update global max with left + right + node.Val

C. Only consider paths from root to leaf nodes

D. Sum all node values regardless of sign and return the total

Solution

1. Step 1: Handle negative child sums carefully

   Use max(0, child sums) to ignore negative paths that reduce the total sum.

2. Step 2: Update global maximum path sum correctly

   Update the global max with the sum of left + right + current node value to consider paths passing through the node.

3. Final Answer:

   Use max(0, child sums) to ignore negative paths and update global max with left + right + node.Val -> Option B

4. Quick Check:

   Ignore negatives, update max with left+right+node [OK]

Hint: Ignore negative child sums, update max with combined path [OK]

Common Mistakes:

• Summing all nodes including negatives blindly
• Considering only root-to-leaf paths
• Removing negative nodes instead of ignoring sums