https://leetcode.com/problems/binary-tree-maximum-path-sum/

本題的難點在於每條路徑可以由樹中的任意兩個節點相連組成，解題方法還是遞迴。需要注意的是遞迴函式的返回值不是子樹的和，而是包含根節點的左子樹、根節點或者包含根節點的右子樹，這也是本題的遞迴函式和其他題目不同的地方。本題的時間複雜度是O(n)，空間複雜度也是O(n)。以下是C++程式碼和python程式碼。
 

class Solution {
public:
	int maxPathSum(TreeNode * root) {
		if (root == NULL)
			return 0;

		int res = root->val;
		helper(root, res);
		return res;
	}

	int helper(TreeNode * root, int & res)
	{
		if (root == NULL)
			return 0;

		int left = helper(root->left, res);
		int right = helper(root->right, res);
		int cur = root->val + (left>0 ? left : 0) + (right>0 ? right : 0);
		if (cur>res)
			res = cur;
		return root->val + max(left, max(right, 0));
	}
};
 

class Solution:
    maxVal = 0

    def maxPathSum(self, root):
        if root == None:
            return 0

        maxVal = root.val
        self.helper(root)
        return maxVal

    def helper(self):
        if root == None:
            return 0

        left = helper(root.left);
        right = helper(root.right);
        cur = root.val + max(left, 0) + max(right, 0)

        if cur > maxVal:
            maxVal = cur

        return root.val + max(left, max(right,0))