We can solve this recursively by processing the tree in reverse pre-order (Right -> Left -> Root). We maintain a prev pointer that points to the previously processed node and attach it to the current node's right.

1. Initialize prev = None.
2. Define a helper flatten(node):
   • If node is None, return.
   • Recurse on the right: flatten(node.right).
   • Recurse on the left: flatten(node.left).
   • Set node.right = prev.
   • Set node.left = None.
   • Update prev = node.
3. The main function calls flatten(root).

• Time Complexity: O(N).
• Space Complexity: O(H).

def flatten(root):
    prev = None
    
    def helper(node):
        nonlocal prev
        if not node:
            return
        
        # Reverse preorder: Right, then Left, then Root
        helper(node.right)
        helper(node.left)
        
        node.right = prev
        node.left = None
        prev = node
        
    helper(root)