使用Python3怎么实现一个三级菜单功能

#!/usr/bin/env python3
# Author:Robert
# --*-- coding: utf-8 --*--
data = {
  "北京":{
    "东城区":{
      "安定门":["国子监大街","孔庙","钟楼"],
      "建国门":["Jinbaojie","长安街","西街"],
      "朝阳门":["东四南大街","朝阳门内大街","孚王府"]
      },
    "朝阳区":{
      "和平街":["胜古庄社区","樱花社区","和平东街社区"],
      "八里庄":["慈寿寺塔","定慧寺"],
      "三里屯":["798艺术区","北京工人体育馆"]
    },
    "海淀":{}
  },
  '山东':{
    "德州":{},
    "青岛":{},
    "济南":{}
  },
  '广东':{
    "东莞":{},
    "常熟":{},
    "佛山":{}
  }
}
exit_flag = False
while not exit_flag:
  for i in data:
    print(i)
  choice = input("选择进入-->:")
  if choice in data:
    while not exit_flag:
      for i2 in data[choice]:
        print("\t",i2)
      choice2 = input("选择进入2-->:")
      if choice2 in data[choice]:
        while not exit_flag:
          for i3 in data[choice][choice2]:
            print("\t\t",i3)
          choice3 = input("选择进入3-->:")
          if choice3 in data[choice][choice2]:
            for i4 in data[choice][choice2][choice3]:
              print("\t\t",i4)
            choice4 = input("最后一层，按b返回-->:")
            if choice4 == "b":
              pass
            elif choice4 == "q":
              exit_flag = True
          if choice3 == "b":
            break
          elif choice4 == "q":
            exit_flag = True
      if choice2 == "b":
        break
      elif choice4 == "q":
        exit_flag = True