Maximum Path Sum in Binary Tree in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to define the TreeNode constructor function.

DSA Javascript

function TreeNode(val) {
  this.val = val;
  this.left = null;
  this.[1] = null;
}

Drag options to blanks, or click blank then click option'

Achild

Bparent

Cnext

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'parent' or 'next' instead of 'right' for the child pointer.

Leaving the property undefined.

The TreeNode has 'left' and 'right' pointers to child nodes. Here, the blank should be 'right' to complete the node structure.

2fill in blank

medium

Complete the code to initialize the maxSum variable to the smallest possible number.

DSA Javascript

let maxSum = [1];

Drag options to blanks, or click blank then click option'

A-Infinity

B0

CNumber.MAX_SAFE_INTEGER

DNumber.MIN_SAFE_INTEGER

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 which may be incorrect if all values are negative.

Using Number.MAX_SAFE_INTEGER which is very large positive number.

We initialize maxSum to -Infinity to ensure any path sum will be larger and update maxSum correctly.

3fill in blank

hard

Fix the error in the recursive function to calculate max path sum from a node.

DSA Javascript

function maxGain(node) {
  if (node === null) return 0;
  let leftGain = Math.max(maxGain(node.left), [1]);
  let rightGain = Math.max(maxGain(node.right), 0);
  let priceNewpath = node.val + leftGain + rightGain;
  maxSum = Math.max(maxSum, priceNewpath);
  return node.val + Math.max(leftGain, rightGain);
}

Drag options to blanks, or click blank then click option'

Anode.val

B0

C-Infinity

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using node.val instead of 0 which is incorrect here.

Using negative infinity which doesn't help avoid negative sums.

We use 0 as the minimum gain from left or right child to avoid negative sums reducing the path.

4fill in blank

hard

Fill both blanks to complete the main function that returns the maximum path sum in the binary tree.

DSA Javascript

var maxPathSum = function(root) {
  maxSum = [1];
  function maxGain(node) {
    if (node === null) return 0;
    let leftGain = Math.max(maxGain(node.left), 0);
    let rightGain = Math.max(maxGain(node.right), 0);
    let priceNewpath = node.val + leftGain + rightGain;
    maxSum = Math.max(maxSum, priceNewpath);
    return node.val + Math.max(leftGain, rightGain);
  }
  maxGain(root);
  return [2];
};

Drag options to blanks, or click blank then click option'

A-Infinity

BmaxSum

C0

Droot.val

Attempts:

3 left

💡 Hint

Common Mistakes

Returning root.val instead of maxSum.

Initializing maxSum to 0 which fails if all values are negative.

Initialize maxSum to -Infinity to handle negative values. Return maxSum after traversal to get the maximum path sum.

5fill in blank

hard

Fill all three blanks to complete the code that calculates the maximum path sum in a binary tree.

DSA Javascript

var maxPathSum = function(root) {
  let maxSum = [1];
  function maxGain(node) {
    if (node === null) return 0;
    let leftGain = Math.max(maxGain(node.[2]), 0);
    let rightGain = Math.max(maxGain(node.[3]), 0);
    let priceNewpath = node.val + leftGain + rightGain;
    maxSum = Math.max(maxSum, priceNewpath);
    return node.val + Math.max(leftGain, rightGain);
  }
  maxGain(root);
  return maxSum;
};

Drag options to blanks, or click blank then click option'

A-Infinity

Bleft

Cright

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'parent' instead of 'left' or 'right' for child nodes.

Initializing maxSum to 0 which fails for all negative nodes.

Initialize maxSum to -Infinity. Use node.left and node.right to traverse children correctly.

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)