Maximum Path Sum in Binary Tree in DSA Java - Build from Scratch

Maximum Path Sum in Binary Tree

📖 Scenario: You are working on a program that analyzes a binary tree representing a network of roads. Each node in the tree has a value representing the road's quality score. You want to find the path through the network that yields the highest total quality score. This path can start and end at any node but must follow parent-child connections.

🎯 Goal: Build a Java program that creates a binary tree with specific node values, then calculates and prints the maximum path sum in the tree.

📋 What You'll Learn

Create a binary tree with given node values using a TreeNode class

Use a helper variable to track the maximum path sum during recursion

Implement a recursive method to compute the maximum path sum

Print the final maximum path sum after processing the tree

💡 Why This Matters

🌍 Real World

Finding the maximum path sum in a binary tree can help in network analysis, such as finding the best route in a road network or optimizing resource flow in hierarchical systems.

💼 Career

This problem is common in technical interviews and helps demonstrate understanding of recursion, tree traversal, and dynamic programming concepts.

1

Create the Binary Tree Structure

Create a class called TreeNode with an int val field, and TreeNode left and TreeNode right fields. Then create a binary tree with root node value 1, left child value 2, and right child value 3.

DSA Java

// Define TreeNode class and create root with left and right children
// Your code here

Hint

Define the TreeNode class with constructor and create the root and its children as specified.

2

Add a Helper Variable for Maximum Path Sum

Inside the Main class, add a static integer variable called maxSum and initialize it to Integer.MIN_VALUE to track the maximum path sum found during recursion.

DSA Java

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

public class Main {
    static int maxSum = Integer.MIN_VALUE; // Your code here
    public static void main(String[] args) {
        TreeNode root = new TreeNode(1);
        root.left = new TreeNode(2);
        root.right = new TreeNode(3);
    }
}

Hint

Add a static variable maxSum in the Main class initialized to Integer.MIN_VALUE.

3

Implement Recursive Method to Find Maximum Path Sum

Add a static method called maxPathSumHelper that takes a TreeNode node parameter and returns an int. This method should recursively compute the maximum path sum starting from the given node. Update the maxSum variable with the maximum path sum found so far. Use the logic: calculate left and right max sums (ignore negatives), update maxSum with the sum of node value plus left and right sums, and return the node value plus the maximum of left or right sums.

DSA Java

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

public class Main {
    static int maxSum = Integer.MIN_VALUE;

    static int maxPathSumHelper(TreeNode node) {
        // Your code here
    }

    public static void main(String[] args) {
        TreeNode root = new TreeNode(1);
        root.left = new TreeNode(2);
        root.right = new TreeNode(3);
    }
}

Hint

Use recursion to get max sums from left and right children, ignore negative sums by comparing with 0, update maxSum, and return the max path sum including current node.

4

Calculate and Print the Maximum Path Sum

In the main method, call maxPathSumHelper with the root node to start the recursion. Then print the value of the maxSum variable.

DSA Java

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

public class Main {
    static int maxSum = Integer.MIN_VALUE;

    static int maxPathSumHelper(TreeNode node) {
        if (node == null) {
            return 0;
        }
        int left = Math.max(0, maxPathSumHelper(node.left));
        int right = Math.max(0, maxPathSumHelper(node.right));
        maxSum = Math.max(maxSum, node.val + left + right);
        return node.val + Math.max(left, right);
    }

    public static void main(String[] args) {
        TreeNode root = new TreeNode(1);
        root.left = new TreeNode(2);
        root.right = new TreeNode(3);

        // Your code here
    }
}

Hint

Call maxPathSumHelper(root) and then print maxSum to display the maximum path sum.

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