<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.maxSum = 0
        def dfs(root):
            if not root:    # If pass the leaf node
                return float("inf"), float("-inf"), 0, True

            # Recursively get to below branches
            min_left, max_left, sum_left, is_bst_left = dfs(root.left)
            min_right, max_right, sum_right, is_bst_right = dfs(root.right)

            # What is BST? --> Contains both left and right BSTs with all elements
            # in left BST must be less than and all elements in right BST must be larger than
            is_BST = is_bst_left and is_bst_right and root.val > max_left and root.val < min_right
            curr_sum = root.val

            if is_BST:
                curr_sum += sum_left + sum_right
                self.maxSum = max(self.maxSum, curr_sum)
            # Continue to compare to BST check on parent nodes since these will be the returned
            # value from left or right branch for its parent node
            curr_min = min(min_left, min_right, root.val)
            curr_max = max(max_left, max_right, root.val)
            return curr_min, curr_max, curr_sum, is_BST

        dfs(root)
        return self.maxSum

        
        
38000