我已经编写了一个测试脚本来评估用户shekhardtu(见答案)和FurkanO(见答案)提出的两种最通用的解决方案的性能(意味着输入不需要事先排序，代码不依赖于第三方库)。

http://playcode.io/316025?tabs=console&script.js&output

FurkanO的解决方案似乎是最快的。

/* ** performance test for https://stackoverflow.com/questions/18017869/build-tree-array-from-flat-array-in-javascript */ // Data Set (e.g. nested comments) var comments = [{ id: 1, parent_id: null }, { id: 2, parent_id: 1 }, { id: 3, parent_id: 4 }, { id: 4, parent_id: null }, { id: 5, parent_id: 4 }]; // add some random entries let maxParentId = 10000; for (let i=6; i<=maxParentId; i++) { let randVal = Math.floor((Math.random() * maxParentId) + 1); comments.push({ id: i, parent_id: (randVal % 200 === 0 ? null : randVal) }); } // solution from user "shekhardtu" (https://stackoverflow.com/a/55241491/5135171) const nest = (items, id = null, link = 'parent_id') => items .filter(item => item[link] === id) .map(item => ({ ...item, children: nest(items, item.id) })); ; // solution from user "FurkanO" (https://stackoverflow.com/a/40732240/5135171) const createDataTree = dataset => { let hashTable = Object.create(null) dataset.forEach( aData => hashTable[aData.id] = { ...aData, children : [] } ) let dataTree = [] dataset.forEach( aData => { if( aData.parent_id ) hashTable[aData.parent_id].children.push(hashTable[aData.id]) else dataTree.push(hashTable[aData.id]) } ) return dataTree }; /* ** lets evaluate the timing for both methods */ let t0 = performance.now(); let createDataTreeResult = createDataTree(comments); let t1 = performance.now(); console.log("Call to createDataTree took " + Math.floor(t1 - t0) + " milliseconds."); t0 = performance.now(); let nestResult = nest(comments); t1 = performance.now(); console.log("Call to nest took " + Math.floor(t1 - t0) + " milliseconds."); //console.log(nestResult); //console.log(createDataTreeResult); // bad, but simple way of comparing object equality console.log(JSON.stringify(nestResult)===JSON.stringify(createDataTreeResult));

我已经编写了一个测试脚本来评估用户shekhardtu(见答案)和FurkanO(见答案)提出的两种最通用的解决方案的性能(意味着输入不需要事先排序，代码不依赖于第三方库)。

http://playcode.io/316025?tabs=console&script.js&output

FurkanO的解决方案似乎是最快的。

/* ** performance test for https://stackoverflow.com/questions/18017869/build-tree-array-from-flat-array-in-javascript */ // Data Set (e.g. nested comments) var comments = [{ id: 1, parent_id: null }, { id: 2, parent_id: 1 }, { id: 3, parent_id: 4 }, { id: 4, parent_id: null }, { id: 5, parent_id: 4 }]; // add some random entries let maxParentId = 10000; for (let i=6; i<=maxParentId; i++) { let randVal = Math.floor((Math.random() * maxParentId) + 1); comments.push({ id: i, parent_id: (randVal % 200 === 0 ? null : randVal) }); } // solution from user "shekhardtu" (https://stackoverflow.com/a/55241491/5135171) const nest = (items, id = null, link = 'parent_id') => items .filter(item => item[link] === id) .map(item => ({ ...item, children: nest(items, item.id) })); ; // solution from user "FurkanO" (https://stackoverflow.com/a/40732240/5135171) const createDataTree = dataset => { let hashTable = Object.create(null) dataset.forEach( aData => hashTable[aData.id] = { ...aData, children : [] } ) let dataTree = [] dataset.forEach( aData => { if( aData.parent_id ) hashTable[aData.parent_id].children.push(hashTable[aData.id]) else dataTree.push(hashTable[aData.id]) } ) return dataTree }; /* ** lets evaluate the timing for both methods */ let t0 = performance.now(); let createDataTreeResult = createDataTree(comments); let t1 = performance.now(); console.log("Call to createDataTree took " + Math.floor(t1 - t0) + " milliseconds."); t0 = performance.now(); let nestResult = nest(comments); t1 = performance.now(); console.log("Call to nest took " + Math.floor(t1 - t0) + " milliseconds."); //console.log(nestResult); //console.log(createDataTreeResult); // bad, but simple way of comparing object equality console.log(JSON.stringify(nestResult)===JSON.stringify(createDataTreeResult));

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);

没有第三方库 不需要预先排序数组 你可以得到树的任何部分

试试这个

function getUnflatten(arr,parentid){
  let output = []
  for(const obj of arr){
    if(obj.parentid == parentid)

      let children = getUnflatten(arr,obj.id)

      if(children.length){
        obj.children = children
      }
      output.push(obj)
    }
  }

  return output
 }

在Jsfiddle上测试它

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。

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/