递归解决用户关系

 public ArrayList getList(ArrayList<QiziUser> list){
	 	ArrayList<ArrayList<QiziUser>> child = new ArrayList<>();
	 	child.add(list);
	 	for(int i=0;i<list.size();i++){
	 		if(!qiziUserService.getTeam(list.get(i).getId()).isEmpty()){
				child.add(getList((ArrayList<QiziUser>) qiziUserService.getTeam(list.get(i).getId())));
			}
	 	}
	 	return child;
	 }

这里就可以把所有用户关系读出来，有点问题，但是已经解决需求，for循环也可以去掉，懒得改了。

//递归获取获取数据
			getTree11(treelist,treelists){
				for(let i=0;i<treelist.length;i++){
					if(treelist[i].constructor===Array){
						this.getTree11(treelist[i],treelists)
					}else{
						const treeNode ={};
						treeNode.id=treelist[i].id;
						treeNode.name =treelist[i].wxName;
						treeNode.userid=treelist[i].userId;
						treeNode.userTouxiang=treelist[i].userTouxiang;
						treeNode.userIntegral=treelist[i].userIntegral;
						if(treelist[i].userMagnitude==0){
						   treeNode.userMagnitude="普通会员"
						}else if(treelist[i].userMagnitude==1){
						    treeNode.userMagnitude="银卡会员"
						}else if(treelist[i].userMagnitude==2){
						     treeNode.userMagnitude="金卡会员"
						}else if(treelist[i].userMagnitude==3){
						     treeNode.userMagnitude="白金会员"
						}else if(treelist[i].userMagnitude==4){
						     treeNode.userMagnitude="铂金会员"
						}else if(treelist[i].userMagnitude==5){
						     treeNode.userMagnitude="钻石会员"
						}else if(treelist[i].userMagnitude==6){
						    treeNode.userMagnitude="股东"
						}else{
							treeNode.userMagnitude="游客"
						}
						treelists.push(treeNode);	
					}
				}
			},

第一次用vue进行递归，也是完成任务了。