<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 maxSumBST(self, root: Optional[TreeNode]) -> int:
        self.mx_sz = 0
        
        n = root
        def f(n):
            if n is None:
                return True, 0, 0, 0
            
            if n.left is None and n.right is None:
                self.mx_sz = max(n.val, self.mx_sz)
                return True, n.val, n.val, n.val
                
            l_res, l_mn, l_mx, l_sz = f(n.left)
            r_res, r_mn, r_mx, r_sz = f(n.right)
            
            if l_res:
                if l_sz > self.mx_sz:
                    self.mx_sz = max(l_sz, self.mx_sz)
                    
            if r_res:
                if r_sz > self.mx_sz:
                    self.mx_sz = max(r_sz, self.mx_sz)
                    
            if n.left is None:
                if r_res and n.val < r_mn:
                    c_res = True
                    c_sz = r_sz + n.val
                    c_mn = n.val
                    c_mx = r_mx
                    self.mx_sz = max(c_sz, self.mx_sz)
                    return True, c_mn, c_mx, c_sz
            elif n.right is None:
                if l_res and l_mx < n.val:
                    c_res = True
                    c_sz = l_sz + n.val
                    c_mn = l_mn
                    c_mx = n.val
                    self.mx_sz = max(c_sz, self.mx_sz)
                    return True, c_mn, c_mx, c_sz
            else:
                if l_res and r_res:
                    # print(n.val, "out", l_sz, r_sz)
                    if l_mx < n.val and n.val < r_mn:
                        c_res = True
                        c_sz = l_sz + r_sz + n.val
                        c_mn = l_mn
                        c_mx = r_mx
                        self.mx_sz = max(c_sz, self.mx_sz)
                        # print(n.val, "here", self.mx_sz)
                        return True, c_mn, c_mx, c_sz
                    
            return False, -1, -1, -1
            
        f(n)
        return self.mx_sz
200