从javascript中的平面数组构建树数组

2022-01-31 00:00:00 list arrays tree javascript

我有一个复杂的 json 文件，我必须用 javascript 处理它以使其分层，以便以后构建一棵树.json的每个条目都有:id :一个唯一的 id，parentId :父节点的id(如果节点是树的根，则为0)level : 树中的深度级别

I have a complex json file that I have to handle with javascript to make it hierarchical, in order to later build a tree. Every entry of the json has : id : a unique id, parentId : the id of the parent node (which is 0 if the node is a root of the tree) level : the level of depth in the tree

json 数据已经有序"了.我的意思是一个条目将在其自身之上有一个父节点或兄弟节点，在其自身之下有一个子节点或兄弟节点.

The json data is already "ordered". I mean that an entry will have above itself a parent node or brother node, and under itself a child node or a brother node.

输入:

{
    "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
        },
    ]
}

预期输出:

{
    "People": [
        {
            "id": "12",
            "parentId": "0",
            "text": "Man",
            "level": "1",
            "children": [
                {
                    "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":
            {

                "id": "11",
                "parentId": "9",
                "text": "Girl",
                "level": "2",
                "children": null
            }
        }

    ],    

    "Animals": [
        {
            "id": "5",
            "parentId": "0",
            "text": "Dog",
            "level": "1",
            "children": 
                {
                    "id": "8",
                    "parentId": "5",
                    "text": "Puppy",
                    "level": "2",
                    "children": null
                }
        },
        {
            "id": "10",
            "parentId": "13",
            "text": "Cat",
            "level": "1",
            "children": 
            {
                "id": "14",
                "parentId": "13",
                "text": "Kitten",
                "level": "2",
                "children": null
            }
        }

    ]
}

推荐答案

如果您使用地图查找，会有一个有效的解决方案.如果父母总是先于他们的孩子，您可以合并两个 for 循环.它支持多个根.它在悬空分支上给出错误，但可以修改为忽略它们.它不需要第 3 方库.据我所知，这是最快的解决方案.

There is an efficient solution if you use a map-lookup. If the parents always come before their children you can merge the two for-loops. It supports multiple roots. It gives an error on dangling branches, but can be modified to ignore them. It doesn't require a 3rd-party library. It's, as far as I can tell, the fastest solution.

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));

如果你研究复杂性理论，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个问题.

If you're into complexity theory this solution is Θ(n log(n)). The recursive-filter solution is Θ(n^2) which can be a problem for large data sets.

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