React Native Navigation, open universal link from within own app

Question

I am using React Navigation for React Native. I have successfully configured it to handle universal link, which is something like this

// linking.ts
import { APP_ID } from '@env';

const config = {
    screens: {
        LoginScreen: 'authorize',
        RegisterScreen: 'register',
        CustomerStack: {
            screens: {
                OrderDetailScreen: 'customer/order/:orderId',
            },
        },
    },
};

const linking = {
    prefixes: [`${APP_ID}://app/`, 'https://example.com/app/'],
    config,
};

export default linking;

// App.tsx

import linking from './linking'

const App = () => {
   return (
      <NavigationContainer linking={linking}> <MyApp /> </NavigationContainer>
   )
}

When I press a link in the browser such as https://example.com/app/customer/order/1234, then it successfully opens my app's order page.

Problem

I want to be able to open the url such as https://example.com/app/customer/order/1234 indside my app and have it open the order page. I have tried

<Button onPress={() => Linking.openURL('https://example.com/app/customer/order/1234')} />

but (testing on IOS) it switch to the web browser first to open the link, and then open my app.

Is it possible to open the order page directly inside my app without switching to the browser first.

Note: I am trying to implement an in-app notification history page, each notification item has the link saved in the database, and when the user clicks on the item I want to navigate the user to the page as configured in linking.ts. I know it is possible to parse the link and use navigation.navigate() instead, but that means I will have 2 places for the linking configuration. I think it would be great if I can reuse the existing logic provided by React Navigation.