[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;
    }
}