/*
    Given root of binary tree, return max path sum (seq of adj node values added together)

    Path can only have <= 1 split point, assume curPath has it, so return can't split again

    Time: O(n)
    Space: O(n)
*/

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int maxPathSum(TreeNode* root) {
        int maxPath = INT_MIN;
        dfs(root, maxPath);
        return maxPath;
    }
private:
    int dfs(TreeNode* root, int& maxPath) {
        if (root == NULL) {
            return 0;
        }
        
        int left = max(dfs(root->left, maxPath), 0);
        int right = max(dfs(root->right, maxPath), 0);
        
        int curPath = root->val + left + right;
        maxPath = max(maxPath, curPath);
        
        return root->val + max(left, right);
    }
};