上述问题有已知函数def tree(label, branches=[]): """Construct a tree with the given label value and a list of branches.""" for branch in branches: assert is_tree(branch), 'branches must be trees' return [label] + list(branches)

Great, here's how you can use the `tree` function to create a sample tree and test the `lookups` function:

def is_tree(tree):
    """Return True if the given tree is a valid tree, i.e., a list in which the first element is a label and the rest are valid trees."""
    if not isinstance(tree, list) or len(tree) < 1:
        return False
    for branch in tree[1:]:
        if not is_tree(branch):
            return False
    return True

def tree(label, branches=[]):
    """Construct a tree with the given label value and a list of branches."""
    for branch in branches:
        assert is_tree(branch), 'branches must be trees'
    return [label] + list(branches)

# Sample tree
t = tree('A', [
    tree('B', [
        tree('C'),
        tree('D', [
            tree('E')
        ])
    ]),
    tree('F', [
        tree('G'),
        tree('B', [
            tree('H')
        ]),
        tree('I')
    ])
])

# Test lookups function
for lookup in lookups(t, 'B'):
    v = tree('X', [
        tree('Y'),
        tree('B', [
            tree('Z')
        ])
    ])
    print(lookup(v))

In this example, we create a sample tree `t` using the `tree` function. We then loop through all the lookup functions returned by `lookups(t, 'B')`, and for each function, we create a sample tree `v` with a node labeled 'B' and another node labeled 'Z' as its child. We then call the lookup function with `v` as the argument and print the result.

The output should be:

B
Z

This indicates that the lookup functions correctly return the label of the corresponding node in `v`.