从平面对象数组构建对象树数组 [复制]

Question

我想从平面数组构建一个树形数组:

下面是平面数组：

nodes = [
    {id: 1, pid: 0, name: "kpittu"},
    {id: 2, pid: 0, name: "news"},
    {id: 3, pid: 0, name: "menu"},
    {id: 4, pid: 3, name: "node"},
    {id: 5, pid: 4, name: "subnode"},
    {id: 6, pid: 1, name: "cace"}
];

NB:id＝节点id；pid=父节点id。

我想将其转换为这个数组：

nodes = [{
    id: 1,
    name: 'kpittu',
    childs: [{
        id: 6,
        name: 'cace'
    }]
}, {
    id: 2,
    name: 'news'
}, {
    id: 3,
    name: 'menu',
    childs: [{
        id: 4,
        name: 'node',
        childs: [{
            id: 5,
            name: 'subnode'
        }]
    }]
}];

我试图使用递归函数来实现预期的结果，但我正在寻找更好的方法。谢谢你的回复。

Answer 1

使用数组#reduce和辅助对象进行迭代：

var nodes = [
  {id: 1, pid: 0, name: "kpittu"},
  {id: 2, pid: 0, name: "news"},
  {id: 3, pid: 0, name: "menu"},
  {id: 4, pid: 3, name: "node"},
  {id: 5, pid: 4, name: "subnode"},
  {id: 6, pid: 1, name: "cace"}
];

const helper = nodes.reduce((h, o) => (h[o.id] = Object.assign({}, o), h), Object.create(null));

const tree = nodes.reduce((t, node) => {
  const current = helper[node.id];
  
  if(current.pid === 0) { // if it doesn't have a parent push to root
    t.push(current);
  } else {
    helper[node.pid].children || (helper[node.pid].children = []) // add the children array to the parent, if it doesn't exist
    helper[node.pid].children.push(current); // push the current item to the parent children array
  }
  
  return t;
}, []);

console.log(tree);

Answer 2

您还可以使用ES6中引入的Map对象。

let nodes = [
  { id: 1, pid: 0, name: "kpittu" },
  { id: 2, pid: 0, name: "news" },
  { id: 3, pid: 0, name: "menu" },
  { id: 4, pid: 3, name: "node" },
  { id: 5, pid: 4, name: "subnode" },
  { id: 6, pid: 1, name: "cace" }
];

function toTree(arr) {
  let arrMap = new Map(arr.map(item => [item.id, item]));
  let tree = [];

  for (let i = 0; i < arr.length; i++) {
    let item = arr[i];

    if (item.pid) {
      let parentItem = arrMap.get(item.pid);

      if (parentItem) {
        let { children } = parentItem;

        if (children) {
          parentItem.children.push(item);
        } else {
          parentItem.children = [item];
        }
      }
    } else {
      tree.push(item);
    }
  }

  return tree;
}

let tree = toTree(nodes);

console.log(tree);

Answer 3

您可以使用散列表，并将每个循环中的< code>id和< code>pid作为连接的节点。

此建议也适用于未排序的数据。

var nodes = [{ id: 6, pid: 1, name: "cace" }, { id: 1, pid: 0, name: "kpittu" }, { id: 2, pid: 0, name: "news" }, { id: 3, pid: 0, name: "menu" }, { id: 4, pid: 3, name: "node" }, { id: 5, pid: 4, name: "subnode" }],
    tree = function (data, root) {
        var r = [], o = {};
        data.forEach(function (a) {
            if (o[a.id] && o[a.id].children) {
                a.children = o[a.id] && o[a.id].children;
            }
            o[a.id] = a;
            if (a.pid === root) {
                r.push(a);
            } else {
                o[a.pid] = o[a.pid] || {};
                o[a.pid].children = o[a.pid].children || [];
                o[a.pid].children.push(a);
            }
        });
        return r;
    }(nodes, 0);

console.log(tree);

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