你可以使用npm包数组到树https://github.com/alferov/array-to-tree。 它将普通的节点数组(带有指向父节点的指针)转换为嵌套的数据结构。

解决了从数据库数据集检索到嵌套数据结构(即导航树)的转换问题。

用法:

var arrayToTree = require('array-to-tree');

var dataOne = [
  {
    id: 1,
    name: 'Portfolio',
    parent_id: undefined
  },
  {
    id: 2,
    name: 'Web Development',
    parent_id: 1
  },
  {
    id: 3,
    name: 'Recent Works',
    parent_id: 2
  },
  {
    id: 4,
    name: 'About Me',
    parent_id: undefined
  }
];

arrayToTree(dataOne);

/*
 * Output:
 *
 * Portfolio
 *   Web Development
 *     Recent Works
 * About Me
 */

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)));

如果使用地图查找，就有一个有效的解决方案。如果父母总是在他们的孩子之前，你可以合并两个for循环。它支持多个根。它在悬垂的分支上给出一个错误，但可以修改为忽略它们。它不需要第三方库。就我所知，这是最快的解决方法。

function list_to_tree(list) { var map = {}, node, roots = [], i; for (i = 0; i < list.length; i += 1) { map[list[i].id] = i; // initialize the map list[i].children = []; // initialize the children } for (i = 0; i < list.length; i += 1) { node = list[i]; if (node.parentId !== "0") { // if you have dangling branches check that map[node.parentId] exists list[map[node.parentId]].children.push(node); } else { roots.push(node); } } return roots; } var entries = [{ "id": "12", "parentId": "0", "text": "Man", "level": "1", "children": null }, { "id": "6", "parentId": "12", "text": "Boy", "level": "2", "children": null }, { "id": "7", "parentId": "12", "text": "Other", "level": "2", "children": null }, { "id": "9", "parentId": "0", "text": "Woman", "level": "1", "children": null }, { "id": "11", "parentId": "9", "text": "Girl", "level": "2", "children": null } ]; console.log(list_to_tree(entries));

如果你喜欢复杂性理论，这个解决方案是Θ(n log(n))。递归过滤器的解决方案是Θ(n^2)，这对于大型数据集可能是一个问题。

这是一个旧线程，但我认为更新永远不会伤害，与ES6你可以做到:

const data = [{ id: 1, parent_id: 0 }, { id: 2, parent_id: 1 }, { id: 3, parent_id: 1 }, { id: 4, parent_id: 2 }, { id: 5, parent_id: 4 }, { id: 8, parent_id: 7 }, { id: 9, parent_id: 8 }, { id: 10, parent_id: 9 }]; const arrayToTree = (items=[], id = null, link = 'parent_id') => items.filter(item => id==null ? !items.some(ele=>ele.id===item[link]) : item[link] === id ).map(item => ({ ...item, children: arrayToTree(items, item.id) })) const temp1=arrayToTree(data) console.log(temp1) const treeToArray = (items=[], key = 'children') => items.reduce((acc, curr) => [...acc, ...treeToArray(curr[key])].map(({ [`${key}`]: child, ...ele }) => ele), items); const temp2=treeToArray(temp1) console.log(temp2)

希望它能帮助到别人

从网上复制 http://jsfiddle.net/stywell/k9x2a3g6/

    function list2tree(data, opt) {
        opt = opt || {};
        var KEY_ID = opt.key_id || 'ID';
        var KEY_PARENT = opt.key_parent || 'FatherID';
        var KEY_CHILD = opt.key_child || 'children';
        var EMPTY_CHILDREN = opt.empty_children;
        var ROOT_ID = opt.root_id || 0;
        var MAP = opt.map || {};
        function getNode(id) {
            var node = []
            for (var i = 0; i < data.length; i++) {
                if (data[i][KEY_PARENT] == id) {
                    for (var k in MAP) {
                        data[i][k] = data[i][MAP[k]];
                    }
                    if (getNode(data[i][KEY_ID]) !== undefined) {
                        data[i][KEY_CHILD] = getNode(data[i][KEY_ID]);
                    } else {
                        if (EMPTY_CHILDREN === null) {
                            data[i][KEY_CHILD] = null;
                        } else if (JSON.stringify(EMPTY_CHILDREN) === '[]') {
                            data[i][KEY_CHILD] = [];
                        }
                    }
                    node.push(data[i]);
                }
            }
            if (node.length == 0) {
                return;
            } else {
                return node;
            }
        }
        return getNode(ROOT_ID)
    }

    var opt = {
        "key_id": "ID",              //节点的ID
        "key_parent": "FatherID",    //节点的父级ID
        "key_child": "children",     //子节点的名称
        "empty_children": [],        //子节点为空时,填充的值  //这个参数为空时,没有子元素的元素不带key_child属性;还可以为null或者[],同理
        "root_id": 0,                //根节点的父级ID
        "map": {                     //在节点内映射一些值  //对象的键是节点的新属性; 对象的值是节点的老属性,会赋值给新属性
            "value": "ID",
            "label": "TypeName",
        }
    };

更新2022

这是一个针对无序项的建议。该函数使用单个循环和哈希表，并收集所有带有id的项。如果找到根节点，则将该对象添加到结果数组中。

const getTree = (data, root) => { const t = {}; data.forEach(o => ((t[o.parentId] ??= {}).children ??= []).push(Object.assign(t[o.id] ??= {}, o))); return t[root].children; }, data = { People: [{ id: "12", parentId: "0", text: "Man", level: "1", children: null }, { id: "6", parentId: "12", text: "Boy", level: "2", children: null }, { id: "7", parentId: "12", text: "Other", level: "2", children: null }, { id: "9", parentId: "0", text: "Woman", level: "1", children: null }, { id: "11", parentId: "9", text: "Girl", level: "2", children: null }], Animals: [{ id: "5", parentId: "0", text: "Dog", level: "1", children: null }, { id: "8", parentId: "5", text: "Puppy", level: "2", children: null }, { id: "10", parentId: "13", text: "Cat", level: "1", children: null }, { id: "14", parentId: "13", text: "Kitten", level: "2", children: null }] }, result = Object.fromEntries(Object .entries(data) .map(([k, v]) => [k, getTree(v, '0')]) ); console.log(result); .as-console-wrapper { max-height: 100% !important; top: 0; }