解析參考：

https://leetcode-cn.com/problems/binary-tree-maximum-path-sum/solution/er-cha-shu-zhong-de-zui-da-lu-jing-he-by-leetcode-/

/**
 * 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 maxval = INT_MIN;
    int maxPathSum(TreeNode* root) {
        if(root==NULL)
            return 0;
        sumtree(root);

        return maxval;
    }
    int sumtree(TreeNode* root){
        if(root==NULL){
            return 0;
        }
        // 貢獻值大於0時，選取該節點
        int
 
 left = max(sumtree(root->left),0);
        int right = max(sumtree(root->right),0);

        maxval =max(maxval, root->val+left+right);  // maxval儲存的是最大值

        return root->val+max(left, right);  // 返回的當前節點的貢獻值

    }
};