It is 1 + max(3, 5) = 6. A downward path chooses one child; it does not visit both branches. If the question asked for total nodes instead, addition would be appropriate.

Represent a binary node as (value, left, right) and an empty tree as None.

def height(node):
    if node is None:
        return 0
    _, left, right = node
    return 1 + max(height(left), height(right))

tree = (8, (3, None, None), (10, None, (12, None, None)))
assert height(tree) == 3
assert height(None) == 0

The leaf 12 returns 1; node 10 receives (0, 1) and returns 2; root 8 receives (1, 2) and returns 3. This is postorder reasoning: finish child dependencies before combining at the parent.

Every node is processed once, so time is O(n) and recursion space is O(h), where h is height. A chain has h=n; a balanced binary tree has logarithmic height. “Tree” alone does not imply logarithmic algorithms.

For a diameter measured in edges, a path through a node could use both child branches, giving left_height + right_height. But the summary sent upward is still the best single downward branch. Sometimes the answer you record at a node differs from what its parent needs.

Analogy limit: a family hierarchy illustrates parent-child structure, but a general graph can have shared descendants and cycles. The assumption that child subtrees are disjoint must be checked before reusing this reasoning.

No. The question needs values and path context, such as the target still remaining after visiting ancestors. The structure stayed the same; the information flow changed.