Maximum Path Sum in Binary Tree in DSA Go - 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 for one side + current node to parent

↓

Repeat for all nodes

↓

Final global max is the answer

We visit each node, find max path sums from left and right children, update global max with paths passing through the node, and return max single-side path sum to parent.

Execution Sample

DSA Go

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

This code finds the maximum path sum in a binary tree by recursively calculating max sums from children and updating a global max.

Execution Table

Step	Operation	Node Visited	Left Max	Right Max	Current Path Sum	Global Max	Return Value	Visual State
1	Visit	Node 1 (10)	?	?	?	-∞	?	10
2	Visit	Node 2 (2)	?	?	?	-∞	?	2
3	Visit	Node 4 (20)	0	0	20	-∞	20	20
4	Return	Node 4	0	0	20	20	20	20
5	Visit	Node 5 (1)	0	0	1	20	1	20 -> 1
6	Return	Node 5	0	0	1	20	1	20 -> 1
7	Calculate	Node 2	20	1	23	23	22	10 -> 2 -> 20, 1
8	Return	Node 2	20	1	23	23	22	10 -> 2 -> 20, 1
9	Visit	Node 3 (-25)	?	?	?	23	?	10 -> 2 -> 20, 1
10	Visit	Node 6 (3)	0	0	3	23	3	10 -> 2 -> 20, 1
11	Return	Node 6	0	0	3	23	3	10 -> 2 -> 20, 1
12	Visit	Node 7 (4)	0	0	4	23	4	10 -> 2 -> 20, 1
13	Return	Node 7	0	0	4	23	4	10 -> 2 -> 20, 1
14	Calculate	Node 3	3	4	-18	23	-21	10 -> 2 -> 20, 1
15	Return	Node 3	3	4	-18	23	-21	10 -> 2 -> 20, 1
16	Calculate	Node 1	22	0	32	32	32	10 -> 2 -> 20, 1, -25 -> 3, 4
17	Return	Node 1	22	0	32	32	32	10 -> 2 -> 20, 1, -25 -> 3, 4
18	End	-	-	-	-	32	-	Final max path sum is 32

💡 All nodes visited, global max updated to 32, recursion ends.

Variable Tracker

Variable	Start	After Step 4	After Step 6	After Step 7	After Step 11	After Step 13	After Step 15	After Step 17	Final
maxSum	-∞	20	20	23	23	23	23	32	32
Return Value at Node 4	N/A	20	20	20	20	20	20	20	20
Return Value at Node 5	N/A	N/A	1	1	1	1	1	1	1
Return Value at Node 2	N/A	N/A	N/A	22	22	22	22	22	22
Return Value at Node 6	N/A	N/A	N/A	N/A	3	3	3	3	3
Return Value at Node 7	N/A	N/A	N/A	N/A	N/A	4	4	4	4
Return Value at Node 3	N/A	N/A	N/A	N/A	N/A	N/A	-21	-21	-21
Return Value at Node 1	N/A	N/A	N/A	N/A	N/A	N/A	N/A	32	32

Key Moments - 3 Insights

Why do we take max(dfs(child), 0) instead of just dfs(child)?

Because negative path sums reduce the total, we ignore them by taking max with 0. See steps 3 and 5 where left and right max are 0 if negative.

Why do we update global max with node.Val + left + right instead of just one side?

Because the max path can pass through the current node connecting left and right subtrees. This is shown in step 7 and step 16 where global max updates consider both sides.

What does the return value represent at each node?

It represents the max path sum for one side plus the current node, to be used by the parent. This is why in step 7 and 16, return values are node.Val + max(left, right).

Visual Quiz - 3 Questions

Look at the execution_table at step 7, what is the global max after visiting Node 2?

A 23

B 20

C 22

D 32

💡 Hint

Check the 'Global Max' column at step 7 in execution_table.

At which step does the recursion return the value 22 for Node 2?

A Step 4

B Step 7

C Step 6

D Step 17

💡 Hint

Look at the 'Return Value' column for Node 2 in execution_table.

If we did not ignore negative sums (did not use max with 0), what would happen to the global max at Node 3 (step 14)?

A It would increase

B It would stay the same

C It would decrease

D It would become zero

💡 Hint

Check how left and right max are 3 and 4, but node.Val is -25, so negative sums affect the total.

Concept Snapshot

Maximum Path Sum in Binary Tree:
- Visit each node recursively
- Calculate max path sum from left and right children (ignore negatives)
- Update global max with node.Val + left + right
- Return max single-side path sum to parent
- Final global max is the answer

Full Transcript

This visualization shows how to find the maximum path sum in a binary tree. We start at the root and recursively visit each node. For each node, we calculate the maximum path sums from its left and right children, ignoring negative sums by taking max with zero. We then calculate the path sum passing through the current node by adding its value and the left and right max sums. We update a global maximum if this path sum is greater. The function returns the maximum path sum for one side plus the current node to its parent. This process repeats for all nodes, and the final global maximum is the maximum path sum in the tree.

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