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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the maximum path sum starting from the current node or 0 if negative.

DSA Java

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

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

Anode.val

B-1

C0

DInteger.MIN_VALUE

Attempts:

3 left

💡 Hint

Common Mistakes

Returning negative values instead of 0 for negative child sums.

Using node.val instead of 0 as the minimum gain.

We use 0 to ignore negative gains from child paths, so maxGain returns at least 0 for negative sums.

2fill in blank

medium

Complete the code to update the global maximum path sum including the current node and both children.

DSA Java

int priceNewpath = node.val + leftGain + [1];
maxSum = Math.max(maxSum, priceNewpath);

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

Anode.val

BmaxSum

C0

DrightGain

Attempts:

3 left

💡 Hint

Common Mistakes

Adding node.val twice.

Using 0 instead of rightGain.

We add rightGain to include the maximum gain from the right child in the new path sum.

3fill in blank

hard

Fix the error in returning the maximum gain to parent node.

DSA Java

return node.val + Math.max(leftGain, [1]);

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ArightGain

BmaxSum

C0

Dnode.val

Attempts:

3 left

💡 Hint

Common Mistakes

Returning maxSum which is a global max, not a gain.

Returning node.val twice.

We return the node value plus the maximum gain from either left or right child to continue the path upwards.

4fill in blank

hard

Fill both blanks to complete the recursive function that calculates maximum path sum.

DSA Java

public int maxPathSum(TreeNode root) {
    maxSum = Integer.MIN_VALUE;
    maxGain(root);
    return [1];
}

private int maxGain(TreeNode node) {
    if (node == null) return 0;
    int leftGain = Math.max(maxGain(node.left), 0);
    int rightGain = Math.max(maxGain(node.right), 0);
    int 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'

AmaxSum

BmaxGain(root)

Cnode.val

DpriceNewpath

Attempts:

3 left

💡 Hint

Common Mistakes

Returning maxGain(root) instead of maxSum.

Returning node.val or priceNewpath directly.

The function returns the global maxSum after computing maxGain from root.

5fill in blank

hard

Fill all three blanks to complete the class with global maxSum and maxGain method.

DSA Java

class Solution {
    private int [1] = Integer.MIN_VALUE;

    public int maxPathSum(TreeNode root) {
        maxGain(root);
        return maxSum;
    }

    private int maxGain(TreeNode node) {
        if (node == null) return 0;
        int leftGain = Math.max(maxGain(node.left), 0);
        int rightGain = Math.max(maxGain(node.right), 0);
        int 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'

AmaxGain

BmaxSum

CmaxPathSum

DpriceNewpath

Attempts:

3 left

💡 Hint

Common Mistakes

Using local variables instead of class fields.

Confusing method names.

maxSum is the global variable, maxGain is the helper method, maxPathSum is the public method.

Practice

1. What does the Maximum Path Sum in Binary Tree problem ask you to find?

easy

A. The highest sum of values along any path in the tree

B. The longest path length in the tree

C. The smallest value in the tree

D. The number of leaf nodes in the tree

Solution

1. Step 1: Understand the problem goal

   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 node but must be connected.

2. Step 2: Differentiate from other tree metrics

   It is not about path length, smallest value, or leaf count. It focuses on sum of node values along a path.

3. Final Answer:

   The highest sum of values along any path in the tree -> Option A

4. Quick Check:

   Maximum path sum = highest sum of connected nodes [OK]

Hint: Focus on sum of node values, not path length or count [OK]

Common Mistakes:

• Confusing path sum with path length
• Thinking path must start at root
• Ignoring negative values in nodes

2. Which of the following Java method signatures is correct for a function that returns the maximum path sum in a binary tree node class TreeNode?

easy

A. public void maxPathSum(TreeNode root) int

B. public int maxPathSum(TreeNode root)

C. int maxPathSum(TreeNode root) void

D. maxPathSum(TreeNode root) -> int

Solution

1. Step 1: Recall Java method syntax

   In Java, a method returning an int with public access and a parameter of type TreeNode is declared as: public int maxPathSum(TreeNode root).

2. Step 2: Check other options for syntax errors

   Options A, B, and D have invalid syntax: missing return type, misplaced void, or arrow notation not used in Java.

3. Final Answer:

   public int maxPathSum(TreeNode root) -> Option B

4. Quick Check:

   Java method syntax = public int methodName(Type param) [OK]

Hint: Remember Java method syntax: access returnType name(params) [OK]

Common Mistakes:

• Using arrow notation like other languages
• Mixing return type and void incorrectly
• Omitting access modifier

3. Given the following binary tree:

    1
   / \
  2   3

What is the maximum path sum?

medium

A. 4

B. 5

C. 3

D. 6

Solution

1. Step 1: Identify all possible paths and their sums

   Paths and sums: 2 alone = 2, 3 alone = 3, 1 alone = 1, 2->1 = 3, 3->1 = 4, 2->1->3 = 6.

2. Step 2: Find the maximum sum

   The maximum sum is 6 from path 2->1->3.

3. Final Answer:

   6 -> Option D

4. Quick Check:

   Sum(2+1+3) = 6 [OK]

Hint: Check sums including root and both children [OK]

Common Mistakes:

• Ignoring paths that go through root connecting both children
• Only considering single branches
• Adding node values incorrectly

4. Consider this Java method snippet for maximum path sum:

int maxPathSum(TreeNode root) {
  if (root == null) return 0;
  int left = maxPathSum(root.left);
  int right = maxPathSum(root.right);
  return Math.max(left, right) + root.val;
}

What is the main issue with this code?

medium

A. It uses Math.max instead of Math.min

B. It returns 0 for null nodes instead of Integer.MIN_VALUE

C. It does not consider paths passing through the root connecting left and right

D. It does not handle negative node values

Solution

1. Step 1: Analyze the recursive return value

   The code returns the max sum of either left or right subtree plus root value, ignoring the case where path passes through root connecting both sides.

2. Step 2: Understand maximum path sum definition

   Maximum path sum can include left subtree + root + right subtree, which this code misses.

3. Final Answer:

   It does not consider paths passing through the root connecting left and right -> Option C

4. Quick Check:

   Missing combined path sum through root [OK]

Hint: Remember max path can include both left and right subtrees [OK]

Common Mistakes:

• Ignoring combined paths through root
• Assuming max of left/right is enough
• Not updating global max path sum

5. You have a binary tree with nodes that can have negative values. Which approach correctly finds the maximum path sum?

hard

A. Use a recursive helper that returns max gain from each node, update global max with max of left + node + right, and ignore negative gains by comparing with 0

B. Sum all node values and return the total as maximum path sum

C. Only consider paths starting from root to leaf nodes

D. Use breadth-first search to find the longest path and sum its nodes

Solution

1. Step 1: Understand handling negative values

   Negative node values can reduce path sum, so ignore negative gains by comparing with 0 when calculating max gain from children.

2. Step 2: Use recursive helper with global max update

   Recursively compute max gain from left and right, update global max with left + node + right, and return max gain for parent calls.

3. Step 3: Reject incorrect approaches

   Summing all nodes ignores path constraints, root-to-leaf only misses other paths, BFS longest path ignores sums.

4. Final Answer:

   Use a recursive helper that returns max gain from each node, update global max with max of left + node + right, and ignore negative gains by comparing with 0 -> Option A

5. Quick Check:

   Recursive max gain + global max update handles negatives [OK]

Hint: Ignore negative gains by max(child, 0) in recursion [OK]

Common Mistakes:

• Summing all nodes regardless of path
• Only considering root-to-leaf paths
• Not handling negative values properly