<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 __init__(self):
        self.min={}
        self.max={}
        self.the={}
    def maxSumBST(self, root: Optional[TreeNode]) -> int:
        # it returns max sum of any BST in the root
        # the method itself visits every node exactly once even without memomization for itself, though it calls self.theBST
        # many times for a single node, which is optimized by memomization
   
        return max(self.maxSumBST(root.left),self.maxSumBST(root.right),self.theBST(root)) if root else 0
    
    def theBST(self,root):
        # it returns max sum of the BST starting from root. if it is not BST, return -float('inf')
        # it visits every node exactly once by memomization
        if root in self.the:
            return self.the[root]
        if root==None: 
            return 0
        left=self.theBST(root.left)
        right=self.theBST(root.right)
        ans=-float('inf')
        
        # this line makes sure that both left and right are valid BST starting from the node and constraint from main root is met
        if left>-float('inf') and right>-float('inf') and self.max_node(root.left)<root.val and self.min_node(root.right)>root.val:
            ans=(left+right+root.val)
        #if root.val==10: print(ans,"!!!!!!")
        self.the[root]=ans
        return ans
        
        
    
    def max_node(self,root):
        # it visits every node exactly once by memomization
        if root in self.max:
            return self.max[root]
        if root==None:
            return -float("inf") # it is counter-intuitive, but it is for self.max_node(None)<root.val
        self.max[root]=max(root.val,self.max_node(root.left),self.max_node(root.right)) # base will not affect real meaning by this line of max
        return self.max[root]
    
    def min_node(self,root):
        # it visits every node exactly once by memomization
        if root in self.min:
            return self.min[root]
        if root==None:
            return float("inf")
        self.min[root]=min(root.val,self.min_node(root.left),self.min_node(root.right))
        return self.min[root]
        
        
    
    
    
        
34000