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Javascript以树状结构排列数组_Javascript_Arrays_Tree_Hierarchy - Fatal编程技术网

Javascript以树状结构排列数组

Javascript以树状结构排列数组,javascript,arrays,tree,hierarchy,Javascript,Arrays,Tree,Hierarchy,我已经在这里创建了一个教程之后的类别层次结构。当我进行查询时,得到的结果与下面的数组转储相同 [ {id:1,name:'A', ancestors:[]} ,{id:2,name:'AA',parent:1, ancestors:[1]} ,{id:4,name:'AAA',parent:2, ancestors:[1,2]} ,{id:5,name:'AAB',parent:2, ancestors:[1,2]}

我已经在这里创建了一个教程之后的类别层次结构。当我进行查询时,得到的结果与下面的数组转储相同

[
    {id:1,name:'A', ancestors:[]}

        ,{id:2,name:'AA',parent:1, ancestors:[1]}

            ,{id:4,name:'AAA',parent:2, ancestors:[1,2]}
            ,{id:5,name:'AAB',parent:2, ancestors:[1,2]}

        ,{id:3,name:'AB',parent:1, ancestors:[1]}

            ,{id:6,name:'ABA',parent:3, ancestors:[1,3]}
            ,{id:7,name:'ABB',parent:3, ancestors:[1,3]}

    ,{id:8,name:'B', ancestors:[]}
]
请帮助我将上述阵列动态转换为以下格式:

[
    {id:1, name:'A', ancestors: [], children : [
        //children of A
        {id:2, name:'AA', parent:1, ancestors:[1], children:[
            //children of AA
            {id:4, name:'AAA', parent:2, ancestors:[1,2], children:[]}
            , {id:5, name:'AAB', parent:2, ancestors:[1,2], children:[]}

        ]}

        , {id:3, name:'AB', parent:1, ancestors:[1], children:[
            //children of AB
            {id:6, name:'ABA', parent:3, ancestors:[1,3], children:[]}
            , {id:7, name:'ABB', parent:3, ancestors:[1,3], children:[]}

        ]}

    ]}

    ,

    {id:8, name:'B', ancestors:[]}
]

到目前为止,我已经能够得出以下结果

通过编写此代码:


下面是我使用递归的解决方案:

function findChildren(array, itemId){
    var newItems = []
    array.forEach(function(el,index){
        if(el.parent == itemId)
            {
                var itemsClone = JSON.parse(JSON.stringify(items));
                itemsClone.pop(index);
                el.children = findChildren(itemsClone, el.id);
                newItems.push(el);
            }


    });
    return newItems;
}


var tree = findChildren(items);

你为什么不选择我的答案而忽略它?
//var items = _GIVEN AT THE TOP_;

var indexedItem = items.map(function(item){
    return item.id;
});

items.forEach(function(item){
    if(!item.parent) {return;}

    var parent = items[indexedItem.indexOf(item.parent)];
    if(!parent.children){parent.children = [];}
    parent.children.push(item.id);
});
function findChildren(array, itemId){
    var newItems = []
    array.forEach(function(el,index){
        if(el.parent == itemId)
            {
                var itemsClone = JSON.parse(JSON.stringify(items));
                itemsClone.pop(index);
                el.children = findChildren(itemsClone, el.id);
                newItems.push(el);
            }


    });
    return newItems;
}


var tree = findChildren(items);