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……