कोडिंग कार्यप्रवाह · बाइनरी सर्च ट्री का सख्त क्रम

def is_strict_bst(root):
    pending, seen, checks = [(root, None, None)], set(), []
    while pending:
        node, low, high = pending.pop()
        if node is None:
            continue
        if type(node) is not list or len(node) != 3 or type(node[0]) is not int:
            raise ValueError("tree.node")
        if id(node) in seen or len(seen) == 1000:
            raise ValueError("tree.structure")
        seen.add(id(node))
        value, left, right = node
        checks.append((value, low, high))
        pending.extend([(right, value, high), (left, low, value)])
    return all((low is None or low < value) and
               (high is None or value < high) for value, low, high in checks)