Devesh is right if the challenge assumes a Binary Search Tree. That tree2 violates BST invariants (left root.val and q.val > root.val:
root = root.right
else:
return root
```

General binary tree LCA (no ordering assumptions):
python
def lca(root, p, q):
if not root or root == p or root == q:
return root
left = lca(root.left, p, q)
right = lca(root.right, p, q)
return root if left and right else left or right


So either update tree2 to be a valid BST, or clarify the challenge is for a general binary tree and don’t use BST-specific logic.