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

我的typescript解决方案，可能对你有帮助:

type ITreeItem<T> = T & {
    children: ITreeItem<T>[],
};

type IItemKey = string | number;

function createTree<T>(
    flatList: T[],
    idKey: IItemKey,
    parentKey: IItemKey,
): ITreeItem<T>[] {
    const tree: ITreeItem<T>[] = [];

    // hash table.
    const mappedArr = {};
    flatList.forEach(el => {
        const elId: IItemKey = el[idKey];

        mappedArr[elId] = el;
        mappedArr[elId].children = [];
    });

    // also you can use Object.values(mappedArr).forEach(...
    // but if you have element which was nested more than one time
    // you should iterate flatList again:
    flatList.forEach((elem: ITreeItem<T>) => {
        const mappedElem = mappedArr[elem[idKey]];

        if (elem[parentKey]) {
            mappedArr[elem[parentKey]].children.push(elem);
        } else {
            tree.push(mappedElem);
        }
    });

    return tree;
}

用法示例:

createTree(yourListData, 'id', 'parentId');

var data = [{"country":"india","gender":"male","type":"lower","class":"X"}, {"country":"china","gender":"female","type":"upper"}, {"country":"india","gender":"female","type":"lower"}, {"country":"india","gender":"female","type":"upper"}]; var seq = ["country","type","gender","class"]; var treeData = createHieArr(data,seq); console.log(treeData) function createHieArr(data,seq){ var hieObj = createHieobj(data,seq,0), hieArr = convertToHieArr(hieObj,"Top Level"); return [{"name": "Top Level", "parent": "null", "children" : hieArr}] function convertToHieArr(eachObj,parent){ var arr = []; for(var i in eachObj){ arr.push({"name":i,"parent":parent,"children":convertToHieArr(eachObj[i],i)}) } return arr; } function createHieobj(data,seq,ind){ var s = seq[ind]; if(s == undefined){ return []; } var childObj = {}; for(var ele of data){ if(ele[s] != undefined){ if(childObj[ele[s]] == undefined){ childObj[ele[s]] = []; } childObj[ele[s]].push(ele); } } ind = ind+1; for(var ch in childObj){ childObj[ch] = createHieobj(childObj[ch],seq,ind) } return childObj; } }

我使用@FurkanO answer并创建了一个可以用于任何对象类型的泛型函数，我还用TypeScript写了这个函数，我更喜欢它，因为它有自动补全功能。

实现:

1. Javascript:

export const flatListToTree = (flatList, idPath, parentIdPath, childListPath, isParent) => {
  const rootParents = [];
  const map = {};
  for (const item of flatList) {
    if (!item[childListPath]) item[childListPath] = [];
    map[item[idPath]] = item;
  }
  for (const item of flatList) {
    const parentId = item[parentIdPath];
    if (isParent(item)) {
      rootParents.push(item);
    } else {
      const parentItem = map[parentId];
      parentItem[childListPath].push(item);
    }
  }
  return rootParents;
};

2. TypeScript:我假设“T”类型有一个属性的孩子列表，你可以改变“childListPath”是一个字符串而不是“keyof T”如果你有不同的用例。

export const flatListToTree = <T>(
  flatList: T[],
  idPath: keyof T,
  parentIdPath: keyof T,
  childListPath: keyof T,
  isParent: (t: T) => boolean,
) => {
  const rootParents: T[] = [];
  const map: any = {};
  for (const item of flatList) {
    if (!(item as any)[childListPath]) (item as any)[childListPath] = [];
    map[item[idPath]] = item;
  }
  for (const item of flatList) {
    const parentId = item[parentIdPath];
    if (isParent(item)) {
      rootParents.push(item);
    } else {
      const parentItem = map[parentId];
      parentItem[childListPath].push(item);
    }
  }
  return rootParents;
};

使用方法:

  const nodes = [
    { id: 2, pid: undefined, children: [] },
    { id: 3, pid: 2 },
    { id: 4, pid: 2 },
    { id: 5, pid: 4 },
    { id: 6, pid: 5 },
    { id: 7, pid: undefined },
    { id: 8, pid: 7 },
  ];
  
  const result = flatListToTree(nodes, "id", "pid", "children", node => node.pid === undefined);

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/

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"
         }
      ]
   }
]```