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/

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。

类似问题的答案:

https://stackoverflow.com/a/61575152/7388356

更新

你可以使用ES6中引入的Map对象。基本上，你不再通过遍历数组来寻找父元素，你只需要从数组中通过父元素的id来获取父元素就像你在数组中通过索引来获取元素一样。

下面是一个简单的例子:

const 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
  }
];

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.parentId !== "0") {
      let parentItem = arrMap.get(item.parentId);

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

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

  return tree;
}

let tree = toTree(people);

console.log(tree);

下面是Steven Harris的一个修改版本，它是普通的ES5，返回一个以id为键的对象，而不是返回顶层和子层的节点数组。

unflattenToObject = function(array, parent) {
  var tree = {};
  parent = typeof parent !== 'undefined' ? parent : {id: 0};

  var childrenArray = array.filter(function(child) {
    return child.parentid == parent.id;
  });

  if (childrenArray.length > 0) {
    var childrenObject = {};
    // Transform children into a hash/object keyed on token
    childrenArray.forEach(function(child) {
      childrenObject[child.id] = child;
    });
    if (parent.id == 0) {
      tree = childrenObject;
    } else {
      parent['children'] = childrenObject;
    }
    childrenArray.forEach(function(child) {
      unflattenToObject(array, child);
    })
  }

  return tree;
};

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}
];
tree = unflattenToObject(arr);

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; }

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);