对于与树相关的问题，我该怎么办？

Question

问题是：定义一个函数bud_leaves，该函数接受树t和值列表val。它会产生一棵与t相同的新树，但是每个旧叶节点都有一个新分支，每个分支对应vals

def sprout_leaves(t, vals):
    """Sprout new leaves containing the data in vals at each leaf in
    the original tree t and return the resulting tree.

    >>> t1 = tree(1, [tree(2), tree(3)])
    >>> print_tree(t1)
    1
      2
      3
    >>> new1 = sprout_leaves(t1, [4, 5])
    >>> print_tree(new1)
    1
      2
        4
        5
      3
        4
        5

    >>> t2 = tree(1, [tree(2, [tree(3)])])
    >>> print_tree(t2)
    1
      2
        3
    >>> new2 = sprout_leaves(t2, [6, 1, 2])
    >>> print_tree(new2)
    1
      2
        3
          6
          1
          2
    """

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)

def label(tree):
    """Return the label value of a tree."""
    return tree[0]

def branches(tree):
    """Return the list of branches of the given tree."""
    return tree[1:]

def is_tree(tree):
    """Returns True if the given tree is a tree, and False otherwise."""
    if type(tree) != list or len(tree) < 1:
        return False
    for branch in branches(tree):
        if not is_tree(branch):
            return False
    return True

def is_leaf(tree):
    """Returns True if the given tree's list of branches is empty, and False
    otherwise.
    """
    return not branches(tree)

What I've tried so far is:
def sprout_leaves(t, vals):
    if is_leaf(t):
        t += [vals]
    else:
        new_branches = []
        for branch in branches(t):
            if is_leaf(label(branch)):
                for j in range(0, len(label(branch))):
                    [label(branch)[j]] += [vals]
                    new_branches += [label(branch)[j]]
            **????**
    return t

我一直坚持用修改后的分支替换树的原始分支。 例如，如果原始树是[1，[2，3]]，而修改后的分支是[2，[4,5]，3，[4，5]]，我应该怎么做才能重新放置[2， [4,5]，3，[4、5]]？ 任何帮助将不胜感激！