python 扁平列表转树状字典

1.城市扁平列表转树状

理解值引用和位置引用，实现一次遍历生成结果，时间复杂度为O(n)

1.1

输入：

area_list = [{"code": "1", "parent_code": "0"},
             {"code": "11", "parent_code": "1"},
             {"code": "111", "parent_code": "11"},
             {"code": "1111", "parent_code": "111"},
             {"code": "11111", "parent_code": "1111"},
             {"code": "111111", "parent_code": "111111"},
             {"code": "2", "parent_code": "0"},
             {"code": "22", "parent_code": "2"},
             {"code": "222", "parent_code": "22"}]

输出：

result = {
    "sub":[
        {
            "code":"1",
            "sub":[
                {
                    "code":"11",
                    "sub":[
                        {
                            "code":"111",
                            "sub":[
                                {
                                    "code":"1111",
                                    "sub":[
                                        {
                                            "code":"11111",
                                            "sub":[

                                            ],
                                            "parent_code":"1111"
                                        }
                                    ],
                                    "parent_code":"111"
                                }
                            ],
                            "parent_code":"11"
                        }
                    ],
                    "parent_code":"1"
                }
            ],
            "parent_code":"0"
        },
        {
            "code":"2",
            "sub":[
                {
                    "code":"22",
                    "sub":[
                        {
                            "code":"222",
                            "sub":[

                            ],
                            "parent_code":"22"
                        }
                    ],
                    "parent_code":"2"
                }
            ],
            "parent_code":"0"
        }
    ]
}

解决方案：

def solution(lists):
    result = {}
    for a in lists:
        a.setdefault("parent_code")
        result.setdefault(a['code'], {"sub": []})
        result.setdefault(a['parent_code'], {"sub": []})
        result[a['code']].update(a)
        result[a['parent_code']]['sub'].append(result[a['code']])
    return result["0"]

1.2

输入：

industry_list = [
    {
        "pid": "女装",
        "id": "连衣裙"
    },
    {
        "id": "女装"
    },
    {
        "pid": "女装",
        "id": "半身裙"
    },
    {
        "pid": "女装",
        "id": "A字裙"
    },
    {
        "id": "数码"
    },
    {
        "pid": "数码",
        "id": "电脑配件"
    },
    {
        "pid": "电脑配件",
        "id": "内存"
    },
]

输出：

result = {
    "数码": {
        "电脑配件": {
            "内存": {}
        }
    },
    "女装": {
        "连衣裙": {},
        "半身裙": {},
        "A字裙": {}
    }
}

解决方案

def solution(lists):
    data2 = {}
    for a in lists:
        a.setdefault("pid", "")
        data2.setdefault(a['id'], {})
        data2.setdefault(a['pid'], {})
        data2[a['pid']][a['id']] = data2[a['id']]
    return data2[""]