An inorder traversal of a valid BST is sorted. If two nodes are swapped, there will be one or two locations where curr.val < prev.val. We find these "dropped" values and swap them back.

1. Initialize first = None, second = None, and prev = TreeNode(-infinity).
2. Perform an inorder traversal (Recursive or Iterative).
3. In each step:
   • If curr.val < prev.val:
     • If first is None: first = prev (the first problematic node).
     • second = curr (the second problematic node).
   • prev = curr.
4. After traversal, swap values: first.val, second.val = second.val, first.val.

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

def recover_tree(root):
    first = second = prev = None
    
    def inorder(node):
        nonlocal first, second, prev
        if not node: return
        
        inorder(node.left)
        
        # Inorder property violation
        if prev and node.val < prev.val:
            if not first:
                first = prev
            second = node
        prev = node
        
        inorder(node.right)
        
    inorder(root)
    # Swap the values
    if first and second:
        first.val, second.val = second.val, first.val