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');

(奖励1:节点可以排序，也可以不排序)

(bonus2:不需要第三方库，纯js)

(BONUS3:用户“Elias Rabl”说这是最高效的解决方案，见下面他的回答)

下面就是:

const createDataTree = dataset => {
  const hashTable = Object.create(null);
  dataset.forEach(aData => hashTable[aData.ID] = {...aData, childNodes: []});
  const dataTree = [];
  dataset.forEach(aData => {
    if(aData.parentID) hashTable[aData.parentID].childNodes.push(hashTable[aData.ID])
    else dataTree.push(hashTable[aData.ID])
  });
  return dataTree;
};

下面是一个测试，它可能会帮助你理解解决方案是如何工作的:

it('creates a correct shape of dataTree', () => {
  const dataSet = [{
    "ID": 1,
    "Phone": "(403) 125-2552",
    "City": "Coevorden",
    "Name": "Grady"
  }, {
    "ID": 2,
    "parentID": 1,
    "Phone": "(979) 486-1932",
    "City": "Chełm",
    "Name": "Scarlet"
  }];

  const expectedDataTree = [{
    "ID": 1,
    "Phone": "(403) 125-2552",
    "City": "Coevorden",
    "Name": "Grady",
    childNodes: [{
      "ID": 2,
      "parentID": 1,
      "Phone": "(979) 486-1932",
      "City": "Chełm",
      "Name": "Scarlet",
      childNodes : []
    }]
  }];

  expect(createDataTree(dataSet)).toEqual(expectedDataTree);
});

经过多次尝试，我得出了这个结论:

const arrayToTree = (arr, parent = 0) => arr .filter(item => item.parent === parent).map(child => ({ ...child, children: arrayToTree(arr, child.index) }));

   

const entries = [ { index: 1, parent: 0 }, { index: 2, parent: 1 }, { index: 3, parent: 2 }, { index: 4, parent: 2 }, { index: 5, parent: 4 }, { index: 6, parent: 5 }, { index: 7, parent: 6 }, { index: 8, parent: 7 }, { index: 9, parent: 8 }, { index: 10, parent: 9 }, { index: 11, parent: 7 }, { index: 13, parent: 11 }, { index: 12, parent: 0 } ]; const arrayToTree = (arr, parent = 0) => arr .filter(item => item.parent === parent) .map(child => ({ ...child, children: arrayToTree(arr, child.index) })); console.log(arrayToTree(entries));

以防有家长需要。参考id 2，它有多个父元素

const dataSet = [{ "ID": 1, "Phone": "(403) 125-2552", "City": "Coevorden", "Name": "Grady" }, {"ID": 2, "Phone": "(403) 125-2552", "City": "Coevorden", "Name": "Grady" }, { "ID": 3, "parentID": [1,2], "Phone": "(979) 486-1932", "City": "Chełm", "Name": "Scarlet" }]; const expectedDataTree = [ { "ID":1, "Phone":"(403) 125-2552", "City":"Coevorden", "Name":"Grady", "childNodes":[{ "ID":2, "parentID":[1,3], "Phone":"(979) 486-1932", "City":"Chełm", "Name":"Scarlet", "childNodes":[] }] }, { "ID":3, "parentID":[], "Phone":"(403) 125-2552", "City":"Coevorden", "Name":"Grady", "childNodes":[ { "ID":2, "parentID":[1,3], "Phone":"(979) 486-1932", "City":"Chełm", "Name":"Scarlet", "childNodes":[] } ] } ]; const createDataTree = dataset => { const hashTable = Object.create(null); dataset.forEach(aData => hashTable[aData.ID] = {...aData, childNodes: []}); const dataTree = []; dataset.forEach(Datae => { if (Datae.parentID && Datae.parentID.length > 0) { Datae.parentID.forEach( aData => { hashTable[aData].childNodes.push(hashTable[Datae.ID]) }); } else{ dataTree.push(hashTable[Datae.ID]) } }); return dataTree; }; window.alert(JSON.stringify(createDataTree(dataSet)));

正如@Sander提到的，@Halcyon的答案假设一个预先排序的数组，下面的不是。(然而，它假设你已经加载了underscore.js -尽管它可以用香草javascript编写):

Code

// Example usage var arr = [ {'id':1 ,'parentid' : 0}, {'id':2 ,'parentid' : 1}, {'id':3 ,'parentid' : 1}, {'id':4 ,'parentid' : 2}, {'id':5 ,'parentid' : 0}, {'id':6 ,'parentid' : 0}, {'id':7 ,'parentid' : 4} ]; unflatten = function( array, parent, tree ){ tree = typeof tree !== 'undefined' ? tree : []; parent = typeof parent !== 'undefined' ? parent : { id: 0 }; var children = _.filter( array, function(child){ return child.parentid == parent.id; }); if( !_.isEmpty( children ) ){ if( parent.id == 0 ){ tree = children; }else{ parent['children'] = children } _.each( children, function( child ){ unflatten( array, child ) } ); } return tree; } tree = unflatten( arr ); document.body.innerHTML = "<pre>" + (JSON.stringify(tree, null, " ")) <script src="https://cdnjs.cloudflare.com/ajax/libs/underscore.js/1.9.1/underscore-min.js"></script>

需求

它假设属性'id'和'parentid'分别表示id和父id。必须有父ID为0的元素，否则将返回一个空数组。孤儿元素及其后代“丢失”

http://jsfiddle.net/LkkwH/1/

Based on @FurkanO's answer, I created another version that does not mutate the origial data (like @Dac0d3r requested). I really liked @shekhardtu's answer, but realized it had to filter through the data many times. I thought a solution could be to use FurkanO's answer by copying the data first. I tried my version in jsperf, and the results where unfortunately (very) bleak... It seems like the accepted answer is really a good one! My version is quite configurable and failsafe though, so I share it with you guys anyway; here is my contribution:

function unflat(data, options = {}) {
    const { id, parentId, childrenKey } = {
        id: "id",
        parentId: "parentId",
        childrenKey: "children",
        ...options
    };
    const copiesById = data.reduce(
        (copies, datum) => ((copies[datum[id]] = datum) && copies),
        {}
    );
    return Object.values(copiesById).reduce(
        (root, datum) => {
            if ( datum[parentId] && copiesById[datum[parentId]] ) {
                copiesById[datum[parentId]][childrenKey] = [ ...copiesById[datum[parentId]][childrenKey], datum ];
            } else {
                root = [ ...root, datum ];
            }
            return root
        }, []
    );
}

const data = [
    {
        "account": "10",
        "name": "Konto 10",
        "parentAccount": null
    },{
        "account": "1010",
        "name": "Konto 1010",
        "parentAccount": "10"
    },{
        "account": "10101",
        "name": "Konto 10101",
        "parentAccount": "1010"
    },{
        "account": "10102",
        "name": "Konto 10102",
        "parentAccount": "1010"
    },{
        "account": "10103",
        "name": "Konto 10103",
        "parentAccount": "1010"
    },{
        "account": "20",
        "name": "Konto 20",
        "parentAccount": null
    },{
        "account": "2020",
        "name": "Konto 2020",
        "parentAccount": "20"
    },{
        "account": "20201",
        "name": "Konto 20201",
        "parentAccount": "2020"
    },{
        "account": "20202",
        "name": "Konto 20202",
        "parentAccount": "2020"
    }
];

const options = {
    id: "account",
    parentId: "parentAccount",
    childrenKey: "children"
};

console.log(
    "Hierarchical tree",
    unflat(data, options)
);

通过options参数，可以配置将哪个属性用作id或父id。也可以配置children属性的名称，如果有人想要“childNodes”:[]或其他什么。

OP可以简单地使用默认选项:

input.People = unflat(input.People);

如果父对象id是假的(null, undefined或其他假的值)或父对象不存在，我们认为该对象是根节点。