我喜欢@WilliamLeung的纯JavaScript解决方案，但有时你需要在现有数组中进行更改，以保持对对象的引用。

function listToTree(data, options) {
  options = options || {};
  var ID_KEY = options.idKey || 'id';
  var PARENT_KEY = options.parentKey || 'parent';
  var CHILDREN_KEY = options.childrenKey || 'children';

  var item, id, parentId;
  var map = {};
    for(var i = 0; i < data.length; i++ ) { // make cache
    if(data[i][ID_KEY]){
      map[data[i][ID_KEY]] = data[i];
      data[i][CHILDREN_KEY] = [];
    }
  }
  for (var i = 0; i < data.length; i++) {
    if(data[i][PARENT_KEY]) { // is a child
      if(map[data[i][PARENT_KEY]]) // for dirty data
      {
        map[data[i][PARENT_KEY]][CHILDREN_KEY].push(data[i]); // add child to parent
        data.splice( i, 1 ); // remove from root
        i--; // iterator correction
      } else {
        data[i][PARENT_KEY] = 0; // clean dirty data
      }
    }
  };
  return data;
}

Exapmle: https://jsfiddle.net/kqw1qsf0/17/

没有第三方库 不需要预先排序数组 你可以得到树的任何部分

试试这个

function getUnflatten(arr,parentid){
  let output = []
  for(const obj of arr){
    if(obj.parentid == parentid)

      let children = getUnflatten(arr,obj.id)

      if(children.length){
        obj.children = children
      }
      output.push(obj)
    }
  }

  return output
 }

在Jsfiddle上测试它

这是我在一个react项目中使用的

// ListToTree.js
import _filter from 'lodash/filter';
import _map from 'lodash/map';

export default (arr, parentIdKey) => _map(_filter(arr, ar => !ar[parentIdKey]), ar => ({
  ...ar,
  children: _filter(arr, { [parentIdKey]: ar.id }),
}));

用法:

// somewhere.js
import ListToTree from '../Transforms/ListToTree';

const arr = [
   {
      "id":"Bci6XhCLZKPXZMUztm1R",
      "name":"Sith"
   },
   {
      "id":"C3D71CMmASiR6FfDPlEy",
      "name":"Luke",
      "parentCategoryId":"ltatOlEkHdVPf49ACCMc"
   },
   {
      "id":"aS8Ag1BQqxkO6iWBFnsf",
      "name":"Obi Wan",
      "parentCategoryId":"ltatOlEkHdVPf49ACCMc"
   },
   {
      "id":"ltatOlEkHdVPf49ACCMc",
      "name":"Jedi"
   },
   {
      "id":"pw3CNdNhnbuxhPar6nOP",
      "name":"Palpatine",
      "parentCategoryId":"Bci6XhCLZKPXZMUztm1R"
   }
];
const response = ListToTree(arr, 'parentCategoryId');

输出:

[
   {
      "id":"Bci6XhCLZKPXZMUztm1R",
      "name":"Sith",
      "children":[
         {
            "id":"pw3CNdNhnbuxhPar6nOP",
            "name":"Palpatine",
            "parentCategoryId":"Bci6XhCLZKPXZMUztm1R"
         }
      ]
   },
   {
      "id":"ltatOlEkHdVPf49ACCMc",
      "name":"Jedi",
      "children":[
         {
            "id":"C3D71CMmASiR6FfDPlEy",
            "name":"Luke",
            "parentCategoryId":"ltatOlEkHdVPf49ACCMc"
         },
         {
            "id":"aS8Ag1BQqxkO6iWBFnsf",
            "name":"Obi Wan",
            "parentCategoryId":"ltatOlEkHdVPf49ACCMc"
         }
      ]
   }
]```

下面是我根据上面的答案创建的一个简单的帮助函数，为通天塔环境量身定制:

import { isEmpty } from 'lodash'

export default function unflattenEntities(entities, parent = {id: null}, tree = []) {

  let children = entities.filter( entity => entity.parent_id == parent.id)

  if (!isEmpty( children )) {
    if ( parent.id == null ) {
      tree = children
    } else {
      parent['children'] = children
    }
    children.map( child => unflattenEntities( entities, child ) )
  }

  return tree

}

你可以用两行代码来解决这个问题:

_(flatArray).forEach(f=>
           {f.nodes=_(flatArray).filter(g=>g.parentId==f.id).value();});

var resultArray=_(flatArray).filter(f=>f.parentId==null).value();

在线测试(查看浏览器控制台以获得创建的树)

要求:

1-安装lodash 4(一个Javascript库，用于操作对象和集合，使用性能方法=>，就像c#中的Linq)

2-如下所示的flatArray:

    var flatArray=
    [{
      id:1,parentId:null,text:"parent1",nodes:[]
    }
   ,{
      id:2,parentId:null,text:"parent2",nodes:[]
    }
    ,
    {
      id:3,parentId:1,text:"childId3Parent1",nodes:[]
    }
    ,
    {
      id:4,parentId:1,text:"childId4Parent1",nodes:[]
    }
    ,
    {
      id:5,parentId:2,text:"childId5Parent2",nodes:[]
    }
    ,
    {
      id:6,parentId:2,text:"childId6Parent2",nodes:[]
    }
    ,
    {
      id:7,parentId:3,text:"childId7Parent3",nodes:[]
    }
    ,
    {
      id:8,parentId:5,text:"childId8Parent5",nodes:[]
    }];

谢谢Bakhshabadi先生

祝你好运

数组元素可以以混乱的顺序排列

let array = [ { id: 1, data: 'something', parent_id: null, children: [] }, { id: 2, data: 'something', parent_id: 1, children: [] }, { id: 5, data: 'something', parent_id: 4, children: [] }, { id: 4, data: 'something', parent_id: 3, children: [] }, { id: 3, data: 'something', parent_id: null, children: [] }, { id: 6, data: 'something', parent_id: null, children: [] } ] function buildTree(array) { let tree = [] for (let i = 0; i < array.length; i++) { if (array[i].parent_id) { let parent = array.filter(elem => elem.id === array[i].parent_id).pop() parent.children.push(array[i]) } else { tree.push(array[i]) } } return tree } const tree = buildTree(array) console.log(tree); .as-console-wrapper { min-height: 100% }