Write a function that checks if a given binary search tree contains a given value.

Question

Im taking this little test on a website and its telling me that i passed the first test case but not the other two this is what i got
class BinarySearchTree:
Node = collections.namedtuple('Node', ['left', 'right', 'value'])
@staticmethod
def contains(root, value):
if root.value == None:
return False
if root.value == value:
return True
elif root.value > value:
return root.left.value == value
return root.right.value == value
n0 = BinarySearchTree.Node(value=0, left = None, right=None)
n1 = BinarySearchTree.Node(value=1, left=n0, right=None)
n3 = BinarySearchTree.Node(value=3, left=None, right=None)
n2 = BinarySearchTree.Node(value=2, left=n1, right=n3)
print(BinarySearchTree.contains(n2, 3))
OUTPUT: True

Answer 1

You don't seem to be traversing the tree deeply enough. At least that's my opinion, as a python beginner.
If you're not at a value, then you need to keep traversing more branches, until you exhaust your search.

Answer 2

How would I accomplish that it seems that adding more if statements wouldn’t be the best idea to go deeper
Any idea on how I can reach deeper