<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
import concurrent.futures
from functools import partial

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val=val
        self.left=left
        self.right=right

class Solution:
    def maxSumBST(self, root: Optional[TreeNode]) -> int:
        self.max_sum = 0

        def dfs(node):
            if not node:
                return float('inf'), float('-inf'), 0
            
            left_min, left_max, left_sum = dfs(node.left)
            right_min, right_max, right_sum = dfs(node.right)

            if left_max < node.val < right_min:
                curr_sum = left_sum + right_sum + node.val
                self.max_sum = max(self.max_sum, curr_sum)
                return min(left_min, node.val), max(right_max, node.val), curr_sum
            
            return float('-inf'), float('inf'), 0
        
        if not root:
            return 0
        
        with concurrent.futures.ThreadPoolExecutor(max_workers=10) as executor:
            left_future = executor.submit(dfs, root.left)
            right_future = executor.submit(dfs, root.right)
            
            left_result = left_future.result()
            right_result = right_future.result()
        
        # Process the root node
        if left_result[1] < root.val < right_result[0]:
            curr_sum = left_result[2] + right_result[2] + root.val
            self.max_sum = max(self.max_sum, curr_sum)
        
        return self.max_sum
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