Binary Tree Maximum Path Sum

Problem Statement:

A path in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence at most once. Note that the path does not need to pass through the root.

The path sum of a path is the sum of the node’s values in the path.

Given the root of a binary tree, return the maximum path sum of any non-empty path.

Example 1: Input: root = [1,2,3] Output: 6 Explanation: The optimal path is 2 → 1 → 3 with a path sum of 2 + 1 + 3 = 6.

Example 2: Input: root = [-10,9,20,null,null,15,7] Output: 42 Explanation: The optimal path is 15 → 20 → 7 with a path sum of 15 + 20 + 7 = 42.

Constraints:

• The number of nodes in the tree is in the range [1, 3 * 104].
• -1000 <= Node.val <= 1000

Solution:

class Solution {
    // Main method to find the maximum path sum
    public int maxPathSum(TreeNode root) {
        int[] maxSum = {Integer.MIN_VALUE};
        maxPathSum(root, maxSum);
        return maxSum[0];
    }
 
    private int maxPathSum(TreeNode root, int[] maxSum) {
        if (root == null) {
            return 0;
        }
 
        // Recursively get the maximum path sums for the left and right subtrees
        // Only consider positive sums, as negative sums would decrease the overall sum
        int leftSum = Math.max(0, maxPathSum(root.left, maxSum));
        int rightSum = Math.max(0, maxPathSum(root.right, maxSum));
 
        // Calculate the maximum path sum for the current node
        // Option 1: Only the current node's value
        // Option 2: Current node's value plus the maximum of left or right subtree sums
        int temp = Math.max(root.val, root.val + Math.max(leftSum, rightSum));
 
        // Calculate the maximum path sum including both left and right subtrees
        // Option 3: Current node's value plus both left and right subtree sums
        int ans = Math.max(temp, root.val + leftSum + rightSum);
 
        // Update the global maximum path sum if the current path sum is greater
        maxSum[0] = Math.max(maxSum[0], ans);
 
        // Return the maximum path sum that can be extended to the parent node
        return temp;
    }
}