已知一颗树的结点序列以及每个结点的度，构造树的兄弟孩子链表，请给我这道题的代码

假设树的结点用下标 0~n-1 表示，其中 0 号结点为根节点。

class Node:
    def __init__(self, val):
        self.val = val
        self.child = None
        self.sibling = None

def construct_tree(nodes, degrees):
    n = len(nodes)
    degree_cnt = [0] * n
    for i in range(n):
        degree_cnt[degrees[i]] += 1

    # 构造叶子节点
    leaves = []
    for i in range(n):
        if degree_cnt[i] == 0:
            leaves.append(Node(nodes[i]))

    # 构造内部节点
    nodes = [Node(nodes[i]) for i in range(n)]
    for i in range(n):
        if degree_cnt[i] > 0:
            child = None
            for j in range(i+1, n):
                if degree_cnt[j] > 0:
                    if child is None:
                        nodes[i].child = nodes[j]
                    else:
                        child.sibling = nodes[j]
                    child = nodes[j]
                    degree_cnt[i] -= 1
                    degree_cnt[j] -= 1
                    if degree_cnt[i] == 0:
                        break

    # 连接根节点
    root = nodes[0]
    for leaf in leaves:
        leaf.sibling = root.child
        root.child = leaf

    return root

上述代码中，我们先统计每个度数出现的次数，然后先构造叶子节点，再构造内部节点，最后将叶子节点连接到根节点上。具体实现中，我们先创建叶子节点和内部节点的列表，然后用两个指针 `i` 和 `j` 分别指向叶子节点和内部节点，然后按度数从小到大依次连接节点。最后，将叶子节点连接到根节点上即可。