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');

也可以使用lodashjs(v4.x)

function buildTree(arr){
  var a=_.keyBy(arr, 'id')
  return _
   .chain(arr)
   .groupBy('parentId')
   .forEach(function(v,k){ 
     k!='0' && (a[k].children=(a[k].children||[]).concat(v));
   })
   .result('0')
   .value();
}

我的解决方案:

允许双向映射(根到叶，叶到根) 返回所有节点、根节点和叶节点 一次数据传递和非常快的性能 香草Javascript

/**
 * 
 * @param data items array
 * @param idKey item's id key (e.g., item.id)
 * @param parentIdKey item's key that points to parent (e.g., item.parentId)
 * @param noParentValue item's parent value when root (e.g., item.parentId === noParentValue => item is root)
 * @param bidirectional should parent reference be added
 */
function flatToTree(data, idKey, parentIdKey, noParentValue = null, bidirectional = true) {
  const nodes = {}, roots = {}, leaves = {};

  // iterate over all data items
  for (const i of data) {

    // add item as a node and possibly as a leaf
    if (nodes[i[idKey]]) { // already seen this item when child was found first
      // add all of the item's data and found children
      nodes[i[idKey]] = Object.assign(nodes[i[idKey]], i);
    } else { // never seen this item
      // add to the nodes map
      nodes[i[idKey]] = Object.assign({ $children: []}, i);
      // assume it's a leaf for now
      leaves[i[idKey]] = nodes[i[idKey]];
    }

    // put the item as a child in parent item and possibly as a root
    if (i[parentIdKey] !== noParentValue) { // item has a parent
      if (nodes[i[parentIdKey]]) { // parent already exist as a node
        // add as a child
        (nodes[i[parentIdKey]].$children || []).push( nodes[i[idKey]] );
      } else { // parent wasn't seen yet
        // add a "dummy" parent to the nodes map and put the item as its child
        nodes[i[parentIdKey]] = { $children: [ nodes[i[idKey]] ] };
      }
      if (bidirectional) {
        // link to the parent
        nodes[i[idKey]].$parent = nodes[i[parentIdKey]];
      }
      // item is definitely not a leaf
      delete leaves[i[parentIdKey]];
    } else { // this is a root item
      roots[i[idKey]] = nodes[i[idKey]];
    }
  }
  return {roots, nodes, leaves};
}

使用的例子:

const data = [{id: 2, parentId: 0}, {id: 1, parentId: 2} /*, ... */];
const { nodes, roots, leaves } = flatToTree(data, 'id', 'parentId', 0);

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/

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",
        }
    };

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。