python通过递归将字典列表转为tree结构

有时候需要返回树状结构，可以尝试以下思路，不对请指导。

# 字典列表如下    
lists = [
        {"id": 1, "name": "a1", "pid": 0},
        {"id": 2, "name": "a1_2", "pid": 1},
        {"id": 3, "name": "a1_3", "pid": 1},
        {"id": 4, "name": "a1_4", "pid": 1},
        {"id": 5, "name": "a2_1", "pid": 2},
        {"id": 6, "name": "a2_2", "pid": 2},
    ]

我想将他转成这样的树状结构

[
        {
            "id":1,
            "name":"a1",
            "pid":0,
            "sub":[
                {
                    "id":2,
                    "name":"a1_2",
                    "pid":1,
                    "sub":[
                        {
                            "id":5,
                            "name":"a2_1",
                            "pid":2,
                            "sub":[]
                        },
                        {
                            "id":6,
                            "name":"a2_2",
                            "pid":2,
                            "sub":[]
                        }
                    ]
                },
                {
                    "id":3,
                    "name":"a1_3",
                    "pid":1,
                    "sub":[ ]
                },
                {
                    "id":4,
                    "name":"a1_4",
                    "pid":1,
                    "sub":[]
                }
            ]
        }
    ]

于是马上撸代码

    def list_to_tree(index, dict_list, sub_field="sub", pid_field="pid", id_field="id"):
        sub = []
        for x in dict_list:
            if x[pid_field] == index:
                x[sub_field] = list_to_tree(x[id_field], dict_list)
                sub.append(x)
        return sub

print(list_to_tree(0, lists))