Binary Tree Maximum Path Sum

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 * 10<sup>4</sup>].
-1000 <= Node.val <= 1000

Solution:

For each node, find the maximum path sum from left side, right side and the current node => "currentNodeSum"
1. Compare and store the maximum value of "maxResult" and "currentNodeSum"

Return the maximum value of left and right, and the current node

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

  class Solution {

      private int maxSum = Integer.MIN_VALUE;

      private int recursion(TreeNode root) {
          if(root == null) return 0;

          int left  = Math.max(recursion(root.left ), 0);
          int right = Math.max(recursion(root.right), 0);

          int currentNodeSum = root.val + left + right;
          maxSum = Math.max(maxSum, currentNodeSum);

          return Math.max(left,right) + root.val;
      }

      public int maxPathSum(TreeNode root) {
          int res = recursion(root);
          return Math.max(maxSum, res);
      }
  }

Binary Tree Maximum Path Sum | Leetcode 124 | Binary Tree | Dynamic programmin

Solution: