Maximum Path Sum in Binary Tree in DSA Go

Mental Model

Find the highest sum of values along any path in a tree, where the path can start and end at any node.

Analogy: Imagine a mountain range where each peak has a height. You want to find the highest possible trail that can go up and down connecting any peaks, not necessarily starting at the base or ending at the top.

      1
     / \
    2   3
   / \   \
  4  -5   6

Dry Run Walkthrough

Input: Binary tree: 1 -> 2 -> 4, 2 -> -5, 1 -> 3 -> 6

Goal: Find the maximum sum path anywhere in the tree

Step 1: Start at leaf node 4, max path sum from 4 is 4

4 (leaf) -> max path sum = 4

Why: Leaf nodes have max path sum equal to their own value

Step 2: At node -5, max path sum is -5

-5 (leaf) -> max path sum = -5

Why: Negative leaf node max path sum is its own value

Step 3: At node 2, calculate max path sum including children: max(4, 0) + 2 + max(-5, 0) = 6

2 -> left child 4, right child -5 -> max path sum through 2 = 6

Why: Ignore negative child sums, add positive child sums to node value

Step 4: At node 6, max path sum is 6

6 (leaf) -> max path sum = 6

Why: Leaf node max path sum is its own value

Step 5: At node 3, max path sum including child: 3 + max(6, 0) = 9

3 -> right child 6 -> max path sum through 3 = 9

Why: Add positive child max path sum to node value

Step 6: At root 1, max path sum including both children: 1 + max(2's max path sum, 0) + max(3's max path sum, 0) = 1 + 6 + 9 = 16

1 -> left child 2 (6), right child 3 (9) -> max path sum through 1 = 16

Why: Combine left and right max sums plus root for max path crossing root

Step 7: Compare all max path sums found: 4, -5, 6, 9, 16 -> maximum is 16

Maximum path sum in tree = 16

Why: Final answer is the highest sum found anywhere in the tree

Result:

Maximum path sum = 16

Annotated Code

DSA Go

package main

import (
	"fmt"
	"math"
)

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

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

func maxPathSum(root *TreeNode) int {
	maxSum := math.MinInt64

	var dfs func(node *TreeNode) int
	dfs = func(node *TreeNode) int {
		if node == nil {
			return 0
		}

		left := max(dfs(node.Left), 0) // ignore negative sums
		right := max(dfs(node.Right), 0)

		currentSum := node.Val + left + right
		if currentSum > maxSum {
			maxSum = currentSum // update max if current path is better
		}

		return node.Val + max(left, right) // return max path sum including one child
	}

	dfs(root)
	return maxSum
}

func main() {
	root := &TreeNode{Val: 1}
	root.Left = &TreeNode{Val: 2}
	root.Right = &TreeNode{Val: 3}
	root.Left.Left = &TreeNode{Val: 4}
	root.Left.Right = &TreeNode{Val: -5}
	root.Right.Right = &TreeNode{Val: 6}

	result := maxPathSum(root)
	fmt.Println(result)
}

left := max(dfs(node.Left), 0) // ignore negative sums

calculate max path sum from left child, ignore negative sums to avoid reducing total

right := max(dfs(node.Right), 0)

calculate max path sum from right child, ignore negative sums

currentSum := node.Val + left + right

sum path through current node including both children

if currentSum > maxSum { maxSum = currentSum // update max if current path is better }

update global max path sum if current path is larger

return node.Val + max(left, right) // return max path sum including one child

return max sum path continuing upwards through one child

Output

16

Complexity Analysis

Time: O(n) because we visit each node once in the depth-first search

Space: O(h) where h is the height of the tree due to recursion stack

vs Alternative: Naive approach checking all paths would be exponential; this uses DFS with pruning negative sums for efficiency

Edge Cases

Empty tree (root is nil)

Returns minimum integer value or 0 depending on implementation; here returns MinInt64

DSA Go

if node == nil { return 0 }

All negative values in tree

Returns the largest (least negative) single node value as max path sum

DSA Go

left := max(dfs(node.Left), 0) // ignore negative sums

Single node tree

Returns the value of the single node as max path sum

DSA Go

if node == nil { return 0 }

When to Use This Pattern

When asked to find the maximum sum path in a tree that can start and end anywhere, use DFS with tracking max sums including both children and pruning negatives.

Common Mistakes

Mistake: Returning sum of both children upwards instead of only one child
Fix: Return node.Val plus max of left or right child sums to maintain valid path upwards

Mistake: Not ignoring negative sums from children, which lowers total path sum
Fix: Use max(childSum, 0) to ignore negative child sums

Summary

Finds the highest sum of values along any path in a binary tree.

Use when you need the maximum sum path that can start and end at any node in the tree.

The key insight is to track max sums from children ignoring negatives and update global max including both children at each node.

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