No. With the convention that a node is its own ancestor, that target is already their lowest common ancestor. State conventions before deriving base cases.

Use unique integer labels, and assume both target labels exist in the tree. A call returns the discovered target, an already found common ancestor, or None if neither target occurs in its subtree.

def common_ancestor(node, a, b):
    if node is None:
        return None
    value, left, right = node
    if value == a or value == b:
        return value
    left_result = common_ancestor(left, a, b)
    right_result = common_ancestor(right, a, b)
    if left_result is not None and right_result is not None:
        return value
    return left_result if left_result is not None else right_result

tree = (1, (2, (4, None, None), (5, None, None)), (3, None, None))
assert common_ancestor(tree, 4, 5) == 2
assert common_ancestor(tree, 4, 3) == 1
assert common_ancestor(tree, 2, 4) == 2

If both child returns contain evidence, the targets split across the current node. If only one does, both targets or the one partial finding lie on that side, so pass the result upward. Worst-case time is O(n) and stack space O(h).

On a BST with both keys present, ordering offers a shortcut: if both targets are smaller, go left; if both larger, go right; otherwise the current node separates them or equals a target. That takes O(h) time. On a general tree, using numeric comparisons this way would be invalid.

For many queries on one static tree, repeated full traversals may be wasteful. Parent/depth preprocessing and binary lifting are an extension once the one-query contract is understood.

Analogy limit: a partial finding is not proof that both requested targets exist. The existence assumption is doing real work in this compact implementation.

No. It may return the one present target. Verify target presence first, or return both a candidate and target-presence information from the traversal. Handle equal targets consistently with the chosen contract.