如何递归遍历对象数组并筛选生成扁平化新数组？

我有如下嵌套对象数组：

[
  {
    "id": "11ed-a261-0242ac120002",
    "name": "Languages",
    "code": 1,
    "description": "Here you can find teacher for any language. Learn when you want and how you want.",
    "childrenCategories": [
      {
        "id": "a261-0242ac120002",
        "name": "German",
        "code": 7,
        "description": "Here you can find teacher for German. Learn when you want and how you want.",
        "childrenCategories": []
      },
      {
        "id": "a261-0242ac120002",
        "name": "Spanish",
        "code": 8,
        "description": "Here you can find teacher for Spanish. Learn when you want and how you want.",
        "childrenCategories": []
      }
    ]
  },
  {
    "id": "a261-0242ac120002",
    "name": "Sciences",
    "code": 3,
    "description": "It is your online university. Learn whatever you want and become better.",
    "childrenCategories": [
      {
        "id": "6a66915a-3439-11ed-a261-0242ac120002",
        "name": "Mathematics",
        "code": 12,
        "description": "It is your online university. Learn Mathematics and become better.",
        "childrenCategories": []
      },
      {
        "id": "6a669420-3439-11ed-a261-0242ac120002",
        "name": "Physics",
        "code": 13,
        "description": "It is your online university. Learn Physics and become better.",
        "childrenCategories": []
      }
    ]
  }
]

需要生成仅包含name和code字段的扁平化数组：

[
  {"name": "Languages", "code": 1},
  {"name": "German", "code": 7},
  {"name": "Spanish", "code": 8},
  {"name": "Sciences", "code": 3},
  {"name": "Mathematics", "code": 12},
  {"name": "Physics", "code": 13}
]

我已经实现了一个可行方案：

const flatten = (arr) => {
  let result = []
  arr.flatMap((item) => {
    result.push({name: item.name, code: item.code})
    item.childrenCategories.map(child =>
      result.push({name: child.name, code: child.code})
    )
  })
  return result
}

以下是其他几种不同的实现方案：

方案1：递归实现（支持任意层级嵌套） #

如果后续分类可能出现更深层级的嵌套，递归方案能通用处理所有深度的子分类：

const flattenCategories = (categories) => {
  let result = []
  categories.forEach(category => {
    // 添加当前分类的核心字段
    result.push({ name: category.name, code: category.code })
    // 递归处理子分类并合并结果
    if (category.childrenCategories.length > 0) {
      result = result.concat(flattenCategories(category.childrenCategories))
    }
  })
  return result
}

方案2：reduce+concat组合（非递归，适配两层嵌套场景） #

利用reduce累加结果，结合concat一次性合并当前项和所有子项：

const flatten = (arr) => {
  return arr.reduce((acc, item) => {
    return acc.concat(
      { name: item.name, code: item.code },
      item.childrenCategories.map(child => ({ name: child.name, code: child.code }))
    )
  }, [])
}

方案3：flatMap链式调用（简洁写法） #

充分利用flatMap自动扁平化数组的特性，直接返回当前项和子项的数组集合：

const flatten = (arr) => {
  return arr.flatMap(item => [
    { name: item.name, code: item.code },
    ...item.childrenCategories.map(child => ({ name: child.name, code: child.code }))
  ])
}

方案4：生成器函数（迭代式扁平化） #

生成器函数可以逐个输出分类项，适合处理大数据量时降低内存占用：

function* generateFlatCategories(categories) {
  for (const category of categories) {
    yield { name: category.name, code: category.code }
    // 迭代子分类
    yield* generateFlatCategories(category.childrenCategories)
  }
}

// 使用时转换为数组
const flatten = (arr) => Array.from(generateFlatCategories(arr))

内容的提问来源于stack exchange，提问作者Vasyl