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新手上路,【比較難】JS多層遞歸遍歷大佬們有什么好的建議？

JavaScript

千萬里不及你 2019-10-16 10:15:22

小弟初學js，遇到一個難題，關于把一個多層數(shù)據(jù)結構的轉換成treeview可識別的格式問題原數(shù)據(jù)格式為：{"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"]}};期望處理成的格式為：[{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"}]}]}]}]原理就是從structure中層層找child，起始點是root節(jié)點的值（即"body"）每個節(jié)點的child如果不在structure一級節(jié)點中，則認為是葉子節(jié)點想了一天，沒有好辦法，來求助各位前輩

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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);先遍歷每個節(jié)點存在hash表里,然后再遍歷,根據(jù)parent關系找到父節(jié)點.像樓上一樣,data.structure這個也可以理解為一張圖,key是一個頂點,數(shù)組是它的鄰邊集合.非遞歸也可以的