新手上路,【比较难】JS多层递归遍历大佬们有什么好的建议？

Question

小弟初学js，遇到一个难题，关于把一个多层数据结构的转换成treeview可识别的格式问题原数据格式为：{"root":"body","structure":{"body":["head","foot"],"head":["control","tabs"],"control":["c123","c456","c789"],"tabs":["33","44"],"foot":["footLeft","divEnd"],"footLeft":["footLeftContainer"],"footLeftContainer":["add22","ffff","fff"],"divEnd":["divEnd$111"]}};期望处理成的格式为：[{text:"body",nodes:[{text:"head",nodes:[{text:"control",nodes:[{text:"c123"},{text:"c456"},{text:"c789"}]},{text:"tabs",nodes:[{text:"t33"},{text:"t44"}]}]},{text:"foot",nodes:[{text:"footLeft",nodes:[{text:"footLeftContainer",nodes:[{text:"add22"},{text:"ffff"},{text:"fff"}]}]},{text:"divEnd",nodes:[{text:"divEnd$111"}]}]}]}]原理就是从structure中层层找child，起始点是root节点的值（即"body"）每个节点的child如果不在structure一级节点中，则认为是叶子节点想了一天，没有好办法，来求助各位前辈

Answer 1

constdata={
root:'body',
structure:{
body:['head','foot'],
head:['control','tabs'],
control:['c123','c456','c789'],
tabs:['33','44'],
foot:['footLeft','divEnd'],
footLeft:['footLeftContainer'],
footLeftContainer:['add22','ffff','fff'],
divEnd:['divEnd$111'],
},
};
/**
*@param{typeofdata}data
*/
functionconver(data){
constret=[];
constcoll=data.structure;
constroot={text:data.root};
constmap={[data.root]:root};
Object.keys(coll).forEach(key=>{
map[key]=map[key]||{text:key};
coll[key].forEach(item=>{
map[item]=map[item]||{text:item};
});
});
Object.keys(coll).forEach(key=>{
coll[key].forEach(item=>{
if(map[key]){
map[key].nodes=map[key].nodes||[];
map[key].nodes.push(map[item]);
}
});
});
console.log(root);
}
conver(data);
先遍历每个节点存在hash表里,然后再遍历,根据parent关系找到父节点.
像楼上一样,data.structure这个也可以理解为一张图,key是一个顶点,数组是它的邻边集合.非递归也可以的