Solution Explanation: Maximum Sum BST in Binary Tree

This problem asks to find the maximum sum of nodes within a subtree that also forms a valid Binary Search Tree (BST). A simple traversal isn't sufficient because we need to simultaneously check for BST properties and track the maximum sum. The optimal solution uses Depth-First Search (DFS) with a clever return value from the recursive function.

Approach: Post-Order DFS with Tuple Return Value

The core idea is to perform a post-order DFS traversal. For each node, we recursively explore the left and right subtrees. The key is what information we return from the recursive calls: a 4-element tuple. This tuple efficiently summarizes the results of the subtree exploration:

1. bst (boolean): Indicates whether the subtree rooted at the current node is a valid BST (true/false).
2. min (integer): The minimum value in the subtree.
3. max (integer): The maximum value in the subtree.
4. sum (integer): The sum of all node values in the subtree.

The dfs function works as follows:

1. Base Case (Leaf Node or null): If the current node is null (empty subtree), it's considered a valid BST with min = ∞, max = -∞, and sum = 0.

2. Recursive Calls: Recursively calls dfs on the left and right subtrees, obtaining their respective tuples.

3. BST Check and Update: A subtree rooted at the current node is a valid BST only if:

   • Both the left and right subtrees are valid BSTs (lbst and rbst are true).
   • The maximum value in the left subtree is strictly less than the current node's value (lmx < root.val).
   • The minimum value in the right subtree is strictly greater than the current node's value (root.val < rmi).

4. Sum Calculation: If the subtree is a valid BST, calculate its sum (s = ls + rs + root.val). Update the global maximum sum (ans) if necessary.

5. Return Value: Return a tuple containing (bst, min, max, sum). The min and max values are updated to reflect the minimum and maximum values of the current subtree.

Time and Space Complexity Analysis

• Time Complexity: O(N), where N is the number of nodes in the tree. Each node is visited exactly once during the DFS traversal.
• Space Complexity: O(H), where H is the height of the tree. This is due to the recursive call stack in the DFS. In the worst case (a skewed tree), H could be N, resulting in O(N) space complexity.

Code Implementation (Python)

class Solution:
    def maxSumBST(self, root: Optional[TreeNode]) -> int:
        self.ans = 0
        inf = float('inf')  # Use float('inf') for infinity
 
        def dfs(node):
            if not node:
                return True, inf, -inf, 0
 
            lbst, lmin, lmax, lsum = dfs(node.left)
            rbst, rmin, rmax, rsum = dfs(node.right)
 
            if lbst and rbst and lmax < node.val < rmin:
                bst = True
                min_val = min(node.val, lmin)
                max_val = max(node.val, rmax)
                sum_val = lsum + rsum + node.val
                self.ans = max(self.ans, sum_val)
                return bst, min_val, max_val, sum_val
            else:
                return False, 0, 0, 0 #Invalid BST
 
        dfs(root)
        return self.ans

The code in other languages (Java, C++, Go, TypeScript) follows the same logic, just with syntax variations for handling tuples/arrays and infinity. The essential structure of the dfs function remains consistent across all languages.