Beispiel #1
0
    """
    An iterative solution with O(N) time complexity and O(1) space complexity
    """
    node = root

    while node:
        if n1.val > node.val and n2.val > node.val:
            node = node.right
        elif n1.val < node.val and n2.val < node.val:
            node = node.left
        else:
            return node


# -- Testing

from random import randint, seed

if __name__ == "__main__":
    seed(69)
    tree = TreeNode(50)
    for _ in range(25):
        tree.insert(randint(0, 100))
    tree.display()

    print('---')
    node0, node1 = tree.search(41), tree.search(56)
    print(f'Nodes: {node0} {node1}')
    print(f'LCA: {lca(tree, node0, node1)}')
    print(f'LCA: {lowest_common_ancestor(tree, node0, node1)}')
Beispiel #2
0
    if not T1 and not T2:
        return True

    # One empty
    if not T1 or not T2:
        return False

    # Non-equal values
    if T1.val != T2.val:
        return False

    # Otherwise, check that subtrees are equivalent
    return is_equal(T1.left, T2.left) and is_equal(T1.right, T2.right)


# -- Testing

from random import randint, seed

if __name__ == "__main__":
    seed(1209)
    T1 = TreeNode(50)
    for _ in range(50):
        T1.insert(randint(0, 100))
    T1.display()

    T2 = T1.search(42)
    T2.display()

    print(check_subtree(T1, T2))