左树右表如何算法实现

Question

如图我想实现点击左边的部门节点，右边加载该部门的所有员工，包括子部门的所有员工。

如上图，点击人事行政部可以显示该部门人员，但是点击上级部门，就无法显示该部门下属部门的所有员工。
求算法！

Answer 1

我自己用递归方法已经实现了：

    public function getUsersTables($department)
    {
        if($department == 1){
            $users = User::all();
            $users->map(function ($item, $key){
                if ($item->status == 1){
                    return $item->status = '正常';
                }else{
                    return $item->status = '禁用';
                }
            });
            $users->map(function ($item, $key){
                if ($item->confirmed == 1){
                    return $item->confirmed = '已激活';
                }else{
                    return $item->confirmed = '未激活';
                }
            });
            return Datatables::of($users)->make(true);
        }else{
            //获取当前部门的所有员工对象
            $before = User::where('department_id', $department)->get();
            foreach ($before as $item){
                array_push($this->child_users, $item);
            }
            //调用递归方法获取当前部门所有下属部门的ID,并循环获取所有部门ID的员工对象
            foreach ($this->getChildDepartment($department) as $department_id){
                $users = User::where('department_id', $department_id)->get();
                foreach ($users as $user){
                    array_push($this->child_users, $user);
                }
            }
            $temp = collect($this->child_users);
            if (isset($temp)){
                $temp->map(function ($item, $key){
                    if ($item->status == 1){
                        return $item->status = '正常';
                    }else{
                        return $item->status = '禁用';
                    }
                });
                $temp->map(function ($item, $key){
                    if ($item->confirmed ==1){
                        return $item->confirmed = '已激活';
                    }else{
                        return $item->confirmed = '未激活';
                    }
                });
                return Datatables::of($temp)->make(true);
            }
        }
    }
    
    /**
     * 递归方式获取当前被选部门下所有子部门ID
     * @param $parent_id
     * @return array
     */
    public function getChildDepartment($parent_id)
    {
        //获取该部门ID所有下属部门ID
        $child_id = DB::table('departments')->select('id')->where('parent_id', $parent_id)->get();

        $temp = collect($child_id);
        //如果集合为空,则返回子部门数组,否则继续获取当前部门集合中下属部门ID
        if ($temp->isEmpty()){
            return $this->departments;
        }else{
            foreach ($child_id as $id){
                array_push($this->departments, $id->id);
                $this->getChildDepartment($id->id);
            }
        }
        return $this->departments;
    }

性能方面可能还有很多需要调优的地方，等有空再弄吧……

Answer 2

之前用的方法是给每个树节点一个编号..例如父节点是01那么子节点是0101,0102这样,以此类推.查询的时候模糊查询就行了..