TypeScript 中的递归函数 - 父数组

Question

我想创建一个递归函数来接收包含 id 和 parent_id 的对象列表。如果元素的父元素在列表中，我想将其删除并将其添加到父元素。

转换这个：

{
  "id": 180,
  "children": [],
  "parent_id": 195,
  "name": "Object 180"
},
{
  "id": 193,
  "children": [],
  "parent_id": 180,
  "name": "Object 193"
},
{
  "id": 194,
  "children": [],
  "parent_id": 180,
  "name": "Object 194"
}
{
  "id": 199,
  "children": [],
  "parent_id": 187,
  "name": "Object 199"
}
{
  "id": 304,
  "children": [],
  "parent_id": 193,
  "name": "Object 304"
}

到这里：

{
  "id": 180,
  "children": [
    {
      "id": 193,
      "children": [
        {
          "id": 304,
          "children": [],
          "parent_id": 193,
          "name": "Object 304"
         }
      ],
      "parent_id": 180,
      "name": "Object 193"
    },
    {
      "id": 194,
      "children": [],
      "parent_id": 180,
      "name": "Object 194"
    }
  ],
  "parent_id": 195,
  "name": "Object 180"
},
{
  "id": 199,
  "children": [],
  "parent_id": 187,
  "name": "Object 199"
}

有时parent_id为null，父母的等级没有限制。

Answer 1

因为 basarat 的回答没有考虑嵌套超过一层的项目。

这是一个创建具有任意嵌套深度的输出的解决方案：

const listToTree = (input) => {
  const map = new Map(input.map((item) => [item.id, item]));
  
  const output = [];
  for (const item of input) {
    if (map.has(item.parent_id)) {
      map.get(item.parent_id).children.push(map.get(item.id));
    } else {
      output.push(map.get(item.id));
    }
  }
  return output;
};

const input = [
  {
    "id": 180,
    "value": 10,
    "children": [],
    "parent_id": 195,
    "name": "Object 180"
  },
  {
    "id": 193,
    "value": 10,
    "children": [],
    "parent_id": 180,
    "name": "Object 193"
  },
  {
    "id": 194,
    "value": 10,
    "children": [],
    "parent_id": 180,
    "name": "Object 194"
  },
  {
    "id": 199,
    "children": [],
    "parent_id": 187,
    "name": "Object 199"
  },
  {
    "id": 304,
    "value": 10,
    "children": [],
    "parent_id": 193,
    "name": "Object 304"
  },
  {
    "id": 305,
    "value": 10,
    "children": [],
    "parent_id": 194,
    "name": "Object 304"
  }
];

const output = listToTree(input);

console.log(output);

编辑：聚合值

如果您想沿结果树的祖先链聚合值，我建议您之后在单独的函数中执行此操作。这将使您的代码更简洁、更易于测试且更具可读性。

实现取决于您的输入数组是否已排序（子项在父项之前）。如果要处理无序输入，则必须遍历每个祖先链。

function aggregateValue(branch) {
  const children = branch.children || [];
  return children.reduce((sum, child) => sum + aggregateValue(child), branch.value || 0);
}

function aggregateValueAlongBranches(tree) {
  return tree.map((branch) => {
    return {
      ...branch,
      aggregatedValue: aggregateValue(branch),
      children: aggregateValueAlongBranches(branch.children),
    };
  });
}

const input = [
  {
    "id": 180,
    "value": 10,
    "children": [
      {
        "id": 193,
        "value": 10,
        "children": [
          {
            "id": 304,
            "value": 10,
            "children": [],
            "parent_id": 193,
            "name": "Object 304"
          }
        ],
        "parent_id": 180,
        "name": "Object 193"
      },
      {
        "id": 194,
        "value": 10,
        "children": [
          {
            "id": 305,
            "value": 10,
            "children": [],
            "parent_id": 194,
            "name": "Object 304"
          }
        ],
        "parent_id": 180,
        "name": "Object 194"
      }
    ],
    "parent_id": 195,
    "name": "Object 180"
  },
  {
    "id": 199,
    "children": [],
    "parent_id": 187,
    "name": "Object 199"
  }
];
const output = aggregateValueAlongBranches(input);

console.log(output);

Answer 2

您不需要递归函数。只需跟踪您已经看过的项目，如果其中存在父项，请添加到 parent.children 或添加新的根节点。

附上一个完整的解决方案示例。

完整代码

type Item = {
    id: number,
    children: Item[],
    parent_id: number,
    name: string,
}

const items: Item[] = [
    {
        "id": 180,
        "children": [],
        "parent_id": 195,
        "name": "Object 180"
    },
    {
        "id": 193,
        "children": [],
        "parent_id": 180,
        "name": "Object 193"
    },
    {
        "id": 194,
        "children": [],
        "parent_id": 180,
        "name": "Object 194"
    },
    {
        "id": 199,
        "children": [],
        "parent_id": 187,
        "name": "Object 199"
    },
    {
        "id": 304,
        "children": [],
        "parent_id": 193,
        "name": "Object 304"
    }
];


function nest(items:Item[]): Item[] {
  const output: Item[] = [];
  const idToItem = new Map<number,Item>();
  for (let item of items) {
      // Either add to parent. Or create a new root level node
      if (idToItem.has(item.parent_id)) {
          idToItem.get(item.parent_id)!.children.push(item);
      } else {
          idToItem.set(item.id, item);
          output.push(item);
      }
  }
  return output;
}

console.log(nest(items));