[LC] 124. Binary Tree Maximum Path Sum

Given a non-empty binary tree, find the maximum path sum.

For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path must contain at least one node and does not need to go through the root.

Example 1:

Input: [1,2,3]

       1
      / 
     2   3

Output: 6

Example 2:

Input: [-10,9,20,null,null,15,7]

   -10
   / 
  9  20
    /  
   15   7

Output: 42

Time: O(N)

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def maxPathSum(self, root: TreeNode) -> int:
        import sys
        self.res = -sys.maxsize - 1
        self.helper(root, self.res)
        return self.res
    
    def helper(self, root, res):
        if root is None:
            return 0
        left = self.helper(root.left, res)
        right = self.helper(root.right, res)
        if left < 0:
            left = 0
        if right < 0:
            right = 0
        cur_max = root.val + left + right
        if cur_max > self.res:
            # resultcan choose from  both children
            self.res = cur_max
        # return back only choose one path
        return root.val + max(left, right)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    int res = Integer.MIN_VALUE;
    public int maxPathSum(TreeNode root) {
        helper(root);
        return res;
    }
    
    private int helper(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int left = helper(root.left);
        int right = helper(root.right);
        left = left < 0 ? 0 : left;
        right = right < 0 ? 0 : right;
        res = Math.max(res, left + right + root.val);
        return Math.max(left, right) + root.val;
    }
}