<p>Given a <strong>binary tree</strong> <code>root</code>, return <em>the maximum sum of all keys of <strong>any</strong> sub-tree which is also a Binary Search Tree (BST)</em>.</p>

<p>Assume a BST is defined as follows:</p>

<ul>
	<li>The left subtree of a node contains only nodes with keys <strong>less than</strong> the node&#39;s key.</li>
	<li>The right subtree of a node contains only nodes with keys <strong>greater than</strong> the node&#39;s key.</li>
	<li>Both the left and right subtrees must also be binary search trees.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2020/01/30/sample_1_1709.png" style="width: 320px; height: 250px;" /></p>

<pre>
<strong>Input:</strong> root = [1,4,3,2,4,2,5,null,null,null,null,null,null,4,6]
<strong>Output:</strong> 20
<strong>Explanation:</strong> Maximum sum in a valid Binary search tree is obtained in root node with key equal to 3.
</pre>

<p><strong class="example">Example 2:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2020/01/30/sample_2_1709.png" style="width: 134px; height: 180px;" /></p>

<pre>
<strong>Input:</strong> root = [4,3,null,1,2]
<strong>Output:</strong> 2
<strong>Explanation:</strong> Maximum sum in a valid Binary search tree is obtained in a single root node with key equal to 2.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> root = [-4,-2,-5]
<strong>Output:</strong> 0
<strong>Explanation:</strong> All values are negatives. Return an empty BST.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 4 * 10<sup>4</sup>]</code>.</li>
	<li><code>-4 * 10<sup>4</sup> &lt;= Node.val &lt;= 4 * 10<sup>4</sup></code></li>
</ul>

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def __init__(self):
        # 记录 BST 最大节点之和
        self.maxSum = 0
    def maxSumBST(self, root: Optional[TreeNode]) -> int:
        self.findMaxMinSum(root)
        return self.maxSum        
    # def isValidBST(self, root: TreeNode) -> bool:
    #     return self._isValidBST(root, None, None)

    # # 定义：该函数返回 root 为根的子树的所有节点是否满足 max.val > root.val > min.val
    # def _isValidBST(self, root: TreeNode, min: TreeNode, max: TreeNode):
    #     # base case
    #     if root is None:
    #         return True
    #     # 若 root.val 不符合 max 和 min 的限制，说明不是合法 BST
    #     if min is not None and root.val <= min.val:
    #         return False
    #     if max is not None and root.val >= max.val:
    #         return False
    #     # 根据定义，限定左子树的最大值是 root.val，右子树的最小值是 root.val
    #     return self._isValidBST(root.left, min, root) and self._isValidBST(root.right, root, max)
    
    def findMaxMinSum(self,root):
        if root is None:
            return [1, float('inf'), float('-inf'), 0]
        left = self.findMaxMinSum(root.left)
        right = self.findMaxMinSum(root.right)
        res = [0, 0, 0, 0]
        if left[0] == 1 and right[0] == 1 and root.val > left[2] and root.val < right[1]:
            # 以 root 为根的二叉树是 BST
            res[0] = 1
            # 计算以 root 为根的这棵 BST 的最小值
            res[1] = min(left[1], root.val)
            # 计算以 root 为根的这棵 BST 的最大值
            res[2] = max(right[2], root.val)
            # 计算以 root 为根的这棵 BST 所有节点之和
            res[3] = left[3] + right[3] + root.val
            # 更新全局变量
            self.maxSum = max(self.maxSum, res[3])
        else:
            # 以 root 为根的二叉树不是 BST
            res[0] = 0
            # 其他的值都没必要计算了，因为用不到
            
        return res

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