Maximum Path Sum in Binary Tree in DSA Javascript - 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 including current node

↓

Update global max if current path sum is higher

↓

Return max path sum for one side + current node

↓

Repeat for all nodes

↓

Final max path sum stored globally

We visit each node, calculate max path sums from left and right children, update global max path sum including current node, and return max sum for one side to parent.

Execution Sample

DSA Javascript

function maxPathSum(root) {
  let maxSum = -Infinity;
  function dfs(node) {
    if (!node) return 0;
    let left = Math.max(dfs(node.left), 0);
    let right = Math.max(dfs(node.right), 0);
    maxSum = Math.max(maxSum, left + right + node.val);
    return Math.max(left, right) + node.val;
  }
  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 Max Path	Global Max	Return Value	Visual State
1	Visit node	10	N/A	N/A	N/A	-Infinity	N/A	10
2	Visit node	2	N/A	N/A	N/A	-Infinity	N/A	2
3	Calculate left max	2	0	0	2	2	2	10 -> 2
4	Visit node	10	2	N/A	N/A	2	N/A	10
5	Visit node	10	2	10	N/A	2	N/A	10 -> 2, 10
6	Calculate max path	10	2	10	22	22	20	10 -> 2, 10
7	Visit node	-25	N/A	N/A	N/A	22	N/A	10 -> 2, 10
8	Visit node	3	N/A	N/A	N/A	22	N/A	10 -> 2, 10
9	Calculate left max	3	0	0	3	22	3	10 -> 2, 10
10	Visit node	-25	3	N/A	N/A	22	N/A	10 -> 2, 10
11	Visit node	4	N/A	N/A	N/A	22	N/A	10 -> 2, 10
12	Calculate right max	-25	3	4	-18	22	-18	10 -> 2, 10
13	Calculate max path	-25	3	4	-18	22	-18	10 -> 2, 10
14	Return to parent	-25	3	4	-18	22	-18	10 -> 2, 10
15	Calculate max path	10	2	10	22	22	20	10 -> 2, 10
16	Update global max	10	2	10	22	22	20	10 -> 2, 10
17	Return to root	10	2	10	22	22	20	10 -> 2, 10
18	Final max path sum	root	N/A	N/A	N/A	22	N/A	10 -> 2, 10

💡 All nodes visited, global max path sum is 22

Variable Tracker

Variable	Start	After Step 3	After Step 9	After Step 12	After Step 16	Final
maxSum	-Infinity	2	22	22	22	22
left	N/A	0	0	3	2	2
right	N/A	0	0	4	10	10
returnValue	N/A	2	3	-18	20	20

Key Moments - 3 Insights

Why do we use Math.max(dfs(node.left), 0) instead of just dfs(node.left)?

Because negative path sums reduce the total, we ignore them by taking max with 0. See execution_table rows 3 and 9 where left or right max is 0 instead of negative.

Why do we update global max with left + right + node.val?

Because the max 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 of dfs represent?

It represents the max path sum for one side plus current node to be used by parent. It never includes both sides to avoid double counting. See returnValue column in execution_table.

Visual Quiz - 3 Questions

Look at the execution_table at step 6, what is the global max path sum?

A 10

B 12

C 22

D -Infinity

💡 Hint

Check the Global Max column at step 6 in execution_table

At which step does the left max path sum for node with value 3 get calculated?

A Step 9

B Step 3

C Step 12

D Step 16

💡 Hint

Look at the Left Max column and Node Visited column in execution_table

If we remove Math.max(..., 0) and allow negative sums, how would the global max change?

A It would stay the same

B It could decrease because negative paths reduce sums

C It would always increase

D It would cause an error

💡 Hint

Refer to key_moments about ignoring negative sums and execution_table rows where 0 replaces negatives

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 left + right + node value
- Return max path sum for one side + node value
- Final global max is 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 sum from its left and right children, ignoring negative sums by taking the maximum with zero. We then update a global maximum path sum if the sum of left max, right max, and the current node's value is greater than the current global max. The function returns the maximum path sum for one side plus the current node's value to its parent. This process repeats for all nodes, and the final global max 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 sum of all node values in the tree

B. The longest path length in the binary tree

C. The highest sum of values along any path between two nodes in the tree

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

Solution

1. Step 1: Understand the problem definition

   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 nodes, but must be connected.

2. Step 2: Differentiate from other path problems

   This is not about longest path length or total sum of all nodes, but the maximum sum of values on any path.

3. Final Answer:

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

4. Quick Check:

   Maximum path sum = highest sum path [OK]

Hint: Focus on sum of connected nodes, not length or single node [OK]

Common Mistakes:

• Confusing maximum path sum with longest path length
• Thinking path must start at root
• Summing all nodes instead of a path

2. Which of the following is the correct base case return value for a recursive function computing maximum path sum in a binary tree node when the node is null?

easy

A. -Infinity

B. 1

C. null

D. 0

Solution

1. Step 1: Consider the meaning of base case for null node

   When the node is null, the maximum gain from this subtree is 0 (choose not to extend path here).

2. Step 2: Why not others?

   -Infinity works only with Math.max(0, result) but 0 is standard and cleaner. Returning 0 allows proper clamping in parent. null or 1 cause type/math errors.

3. Final Answer:

   0 -> Option D

4. Quick Check:

   Base case for null = 0 [OK]

Hint: Return 0 for null to represent zero contribution [OK]

Common Mistakes:

• Returning -Infinity without Math.max(0, ·)
• Returning null causing type errors
• Returning positive values skewing max

3. What is the output of the following code snippet for maximum path sum?

class TreeNode {
  constructor(val, left = null, right = null) {
    this.val = val;
    this.left = left;
    this.right = right;
  }
}

function maxPathSum(root) {
  let maxSum = -Infinity;
  function dfs(node) {
    if (!node) return 0;
    let left = Math.max(dfs(node.left), 0);
    let right = Math.max(dfs(node.right), 0);
    maxSum = Math.max(maxSum, node.val + left + right);
    return node.val + Math.max(left, right);
  }
  dfs(root);
  return maxSum;
}

const root = new TreeNode(-10, new TreeNode(9), new TreeNode(20, new TreeNode(15), new TreeNode(7)));
console.log(maxPathSum(root));

medium

A. 42

B. 35

C. 27

D. 20

Solution

1. Step 1: Trace dfs on left subtree

   Left child 9: left=0, right=0, maxSum = 9, returns 9.

2. Step 2: Trace dfs on right subtree

   dfs(15)=15 updates maxSum=15; dfs(7)=7 keeps 15; at 20: left=15, right=7, maxSum = max(15, 20+15+7=42)=42, returns 20+15=35.

3. Step 3: Calculate root node max sum

   Root: left=9, right=35, path sum -10+9+35=34 <42, maxSum remains 42.

4. Final Answer:

   42 -> Option A

5. Quick Check:

   Max path sum = 42 [OK]

Hint: Add node + max(left, right) and update maxSum with both sides [OK]

Common Mistakes:

• Not ignoring negative sums by using Math.max with 0
• Adding left and right children incorrectly
• Confusing return value with maxSum update

4. Identify the bug in this maximum path sum function and select the fix:

function maxPathSum(root) {
  let maxSum = -Infinity;
  function dfs(node) {
    if (!node) return -Infinity;
    let left = dfs(node.left);
    let right = dfs(node.right);
    maxSum = Math.max(maxSum, node.val + left + right);
    return node.val + Math.max(left, right);
  }
  dfs(root);
  return maxSum;
}

medium

A. Add check to ignore negative sums with Math.max(left, 0)

B. Change base case return from -Infinity to 0

C. Initialize maxSum to 0 instead of -Infinity

D. Return maxSum inside dfs instead of outside

Solution

1. Step 1: Analyze base case and recursive calls

   Returning -Infinity for null nodes causes left and right to be -Infinity, which breaks addition and max calculations.

2. Step 2: Fix by ignoring negative sums

   Use Math.max(dfs(node.left), 0) and Math.max(dfs(node.right), 0) to ignore negative paths and avoid incorrect sums.

3. Final Answer:

   Add check to ignore negative sums with Math.max(left, 0) -> Option A

4. Quick Check:

   Ignore negative sums with Math.max(..., 0) [OK]

Hint: Ignore negative child sums by max with 0 to fix bugs [OK]

Common Mistakes:

• Returning -Infinity breaks addition
• Not ignoring negative sums from children
• Changing maxSum initialization wrongly

5. Given a binary tree where some nodes have negative values, how would you modify the maximum path sum algorithm to handle the case where all values are negative?

hard

A. Allow paths to be empty and return 0 if all sums are negative

B. Track the maximum single node value separately and compare with path sums

C. Use Math.max with 0 on child sums to ignore negatives

D. Initialize maxSum to -Infinity and do not ignore negative child sums

Solution

1. Step 1: Understand all-negative case

   The max path sum is the highest (least negative) single node value. Paths must include at least one node.

2. Step 2: Key modification

   Tracking the maximum single node value separately ensures that if all child sums are negative, the algorithm still returns the maximum node value. Using Math.max(0, dfs(child)) ignores negative child sums but can miss the max single node if all are negative.

3. Final Answer:

   Track the maximum single node value separately and compare with path sums -> Option B

4. Quick Check:

   Separate max node tracking handles all-negative correctly [OK]

Hint: Track max single node separately to handle all-negative cases [OK]

Common Mistakes:

• Allowing empty path return 0 (invalid)
• Not tracking max node separately (misses max negative node)
• Not ignoring negative child sums (misses single nodes)