<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 State:
    def __init__(self, best_sum: int, cur_sum: Optional[int] = None, value_range: Optional[Tuple[int, int]] = None):
        self.best_sum = best_sum
        self.cur_sum = cur_sum
        self.value_range = value_range

def generate(cur: int, left: Optional[State], right: Optional[State]) -> State:
    if left:
        if right:
            if left.cur_sum is not None and right.cur_sum is not None and left.value_range[1] < cur < right.value_range[0]:
                curSum = left.cur_sum + right.cur_sum + cur
                return State(best_sum = max(curSum, left.best_sum, right.best_sum), cur_sum = curSum, value_range = (left.value_range[0], right.value_range[1]))
            else:
                return State(best_sum = max(left.best_sum, right.best_sum))
        else:
            if left.cur_sum is not None and left.value_range[1] < cur:
                curSum = left.cur_sum + cur
                return State(best_sum = max(curSum, left.best_sum), cur_sum = curSum, value_range = (left.value_range[0], cur))
            else:
                return State(best_sum = left.best_sum)
    else:
        if right:
            if right.cur_sum is not None and right.value_range[0] > cur:
                curSum = right.cur_sum + cur
                return State(best_sum = max(curSum, right.best_sum), cur_sum = curSum, value_range = (cur, right.value_range[1]))
            else:
                return State(best_sum = right.best_sum)
        else:
            return State(best_sum = cur, cur_sum = cur, value_range = (cur, cur))

class Solution:
    def maxSumBST(self, root: Optional[TreeNode]) -> int:
        def search(p: Optional[TreeNode]) -> Optional[State]:
            if not p:
                return None
            return generate(p.val, search(p.left), search(p.right))
        
        return max(search(root).best_sum, 0)
        
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