<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>

#Time: O(N)，因为每个节点被遍历一次。
#Space: O(N)，因为最坏情况下递归调用栈的深度为 N。
class Solution:
    def maxSumBST(self, root: Optional[TreeNode]) -> int:
        result = 0
        
        
        #post order
        def helper(node):
            nonlocal result
            
            if(not node):return True, 0, math.inf, -math.inf  #isBST, subtreeSum, min, max
            
                                #left subtree leftMost val -> leftMin
                                          # #left subtree root val -> leftMax
            isLeftBST, leftSum, leftMin, leftMax = helper(node.left)        #isBST, subtreeSum, min, max
            
                                  #right subtree root val -> rightMin
                                             #right subtree rightMost val -> rightMax
            isRightBST, rightSum, rightMin, rightMax = helper(node.right)   #isBST, subtreeSum, min, max
            
                                             #root 和left/right node 比较
            if(isLeftBST and isRightBST and leftMax < node.val < rightMin):
                subtreeSum = node.val + leftSum + rightSum
                result = max(result, subtreeSum)      
                
                #update完 result之后，还是要return数据上去
                #why min(leftMin, node.val) & max(rightMax, node.val) ?
                    #如果下一层有null child, 就会return True, 0, math.inf, -math.inf  #isBST, subtreeSum, min, max
                        #if(not node):return True, 0, math.inf, -math.inf  #isBST, subtreeSum, min, max
                    # 这里的 isBST是true，但min是inf，max是-inf，所以即使leftMax < node.val < rightMin
                    # 还是要判断min(leftMin, node.val), max(rightMax, node.val)
                return True, subtreeSum, min(leftMin, node.val), max(rightMax, node.val)  #isBST, subtreeSum, min, max
                                                                                    
            else:
                return False, 0, math.inf, -math.inf   #isBST, subtreeSum, min, max  <- reset全部
        
        helper(root)
        return result
213