<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:
    # is_bst, max_ans, total_sum, min_val, max_val
    def check_bst(self, root):
        if(root == None):
            return True, 0, 0, float("inf"), float("-inf")
        
        left = self.check_bst(root.left)
        right = self.check_bst(root.right)
        
        is_bst = (left[0] and right[0] and left[4] < root.val and right[3] > root.val)
        
        max_ans = 0
        max_ans = max(right[1], left[1])
        
        if(is_bst):
            max_ans = max(max_ans, left[2] + right[2] + root.val)

        total_sum = left[2] + right[2] + root.val

        min_val = min(left[3], right[3], root.val)
        max_val = max(left[4], right[4], root.val)
        # print(root.val, is_bst, max_ans, total_sum, min_val, max_val)

        return is_bst, max_ans, total_sum, min_val, max_val




    def maxSumBST(self, root: Optional[TreeNode]) -> int:
        """
        max sum binary tree:
        how can we do it with recursion
        check for left subtree if it is a bst or not
        check for right subtree if it is a bst or not 
        it will return true only if it both are bst and the max of left subtree < root
        min of right subtree < root: two ways: either return or just call again: too much calling and stuff
        use dp?
        also return the sum of all the nodes 
        sum of subtree, max_ans, max_val, min_val, is_bt
        equal is not allowed 
        """

        return self.check_bst(root)[1]
        
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