使用JavaScript构建层次树

3

我正在尝试从我从 SOLR 得到的属性中构建类别、子类别等层次树形结构。输入的格式如下:

['445',
79,
'398',
73,
'710',
32,
'398|760',
28,
'398|760|779',
28,
'445|446',
10]

单引号中的数据表示类别,之后的数字表示频率。

给定上述数组,则需要输出以下格式:

[
    {
        "id": 445,
        "count": 79,
        "children": [
            {
                "id": 446,
                "count": 10
            }
        ]
    },
    {
        "id": 398,
        "count": 73,
        "children": [
            {
                "id": 760,
                "count": 28,
                "children": [
                    {
                        "id": 779,
                        "count": 28
                    }
                ]
            }
        ]
    },
    {
        "id": 710,
        "count": 32
    }
]

我想要用同样的元素构建一棵树,但是一直没有成功。有没有人知道如何让它工作,或者有没有其他时间效率更高的解决方案。

谢谢!


我对这里预期的trie用途感到相当困惑。通常,你会在JavaScript中创建一个trie用于自动完成,比如当用户输入一个类别名称时,而显然你正在处理类别ID...你希望实现什么样的trie结构呢?给定398|730|607,我可以为您设计一个树,例如{398:{freq:73,760:{freq:28,779:{freq:28}},730:{freq:18,607:{freq:18}}}},这是您想要的吗?还是您希望基于类别名称构建一棵树,像这样:{'j':{'a':{'v':{'a':{...}}}}} - Ultimater
我已经编辑了问题,以展示我正在努力实现的最终输出结果,我考虑使用Trie作为字典,然后使用Trie来构建最终的数据结构(而Radix树不仅仅用于自动完成)。但是只要复杂度保持在O(n)的范围内,任何其他方法都可以使用。 - Ajay Pal Singh
为什么不先将数据分割,以便首先获取树的顶点,然后再获取边缘?或者它总是这样的:你首先获取边缘,然后获取顶点? - Christo S. Christov
源数据的来源超出了我的控制,因此我不能假设它,但我可以修改它以符合您提到的格式 - 但我无法理解这将如何有助于减少复杂性。 我通过在桶内递归地形成桶来解决了问题 - 但是很容易预料到,时间和空间方面都会产生可怕的权衡! - Ajay Pal Singh
1个回答

1

这应该可以满足您的需求,但是请注意,没有子节点的节点仍将具有node.children属性,因此您可以使用node.children.length查看树中任何节点的子节点数量。

var o=["445", 79, "398", 73, "710", 32, "398|760", 28, "398|760|779", 28, "445|446", 26, "710|1045", 25, "445|452", 24, "381", 19, "445|943", 19, "398|730", 18, "398|730|607", 18, "367", 16, "445|446|451", 15, "351", 14, "351|363", 14, "351|363|365", 14, "381|395", 14, "381|395|566", 14, "445|526", 14, "445|526|769", 14, "367|372", 12, "710|1045|1119", 11, "398|410", 10, "398|483", 9, "445|452|743", 8, "367|372|377", 7, "398|483|757", 7, "445|446|792", 7, "445|452|744", 7, "445|452|719", 6, "398|410|411", 5];

var nodeMap={};
var nodeLevels=[];
for(var i=0;i<o.length;i+=2)
{
    var catLineage=o[i].split('|');
    var cat=catLineage[catLineage.length-1];
    var depth=catLineage.length;
    while(depth>nodeLevels.length){nodeLevels.push([]);}
    nodeMap[cat]={id:cat,count:o[i+1],depth:depth,parents:catLineage.slice(0,catLineage.length-1)};
    nodeLevels[depth-1].push(cat);
}
var tree=[];
var treeNodeLookup={};
for(var i=0;i<nodeLevels.length;i++)
{
    for(var j=0;j<nodeLevels[i].length;j++)
    {
        var nodeId=nodeLevels[i][j];
        var nodeDepth=nodeMap[nodeId].depth;
        var nodeCount=nodeMap[nodeId].count;
        var parents=nodeMap[nodeId].parents;
        var pointer={children:tree};
        if(parents.length>0){pointer=treeNodeLookup[parents[0]];}
        var node={id:nodeId,count:nodeCount,children:[]};
        pointer.children.push(node);
        treeNodeLookup[nodeId]=pointer.children[pointer.children.length-1];
    }
}
console.log(tree);

使用 console.log(JSON.stringify(tree));,我的输出如下:
[{"id":"445","count":79,"children":[{"id":"446","count":26,"children":[]},{"id":"452","count":24,"children":[]},{"id":"943","count":19,"children":[]},{"id":"526","count":14,"children":[]},{"id":"451","count":15,"children":[]},{"id":"769","count":14,"children":[]},{"id":"743","count":8,"children":[]},{"id":"792","count":7,"children":[]},{"id":"744","count":7,"children":[]},{"id":"719","count":6,"children":[]}]},{"id":"398","count":73,"children":[{"id":"760","count":28,"children":[]},{"id":"730","count":18,"children":[]},{"id":"410","count":10,"children":[]},{"id":"483","count":9,"children":[]},{"id":"779","count":28,"children":[]},{"id":"607","count":18,"children":[]},{"id":"757","count":7,"children":[]},{"id":"411","count":5,"children":[]}]},{"id":"710","count":32,"children":[{"id":"1045","count":25,"children":[]},{"id":"1119","count":11,"children":[]}]},{"id":"381","count":19,"children":[{"id":"395","count":14,"children":[]},{"id":"566","count":14,"children":[]}]},{"id":"367","count":16,"children":[{"id":"372","count":12,"children":[]},{"id":"377","count":7,"children":[]}]},{"id":"351","count":14,"children":[{"id":"363","count":14,"children":[]},{"id":"365","count":14,"children":[]}]}]

非常感谢!看起来完美——空的子元素属性根本不是问题!非常感谢您的帮助!我想这没有办法简化为O(n)! - Ajay Pal Singh
如果您想添加大O复杂度支持,应该使用节点树,这与trie非常不同。我认为像JavaScript这样的高级语言,由许多浏览器解释,或者对于node.js来说是JavaScript引擎,无法理解这种性能计算。但我可以理解,像Java这样的低级语言会因其强类型而受益匪浅。 - Ultimater

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