Create Cluster (WSFC Needed for HA_Alwayson Availability Group) on 2 Separate Datacenters

Hi
I want to run HADR for our servers. (Alwayson - Availability Group )
If I got it right , Disaster server (secondary replica for disaster recovery) Should be install on another site, means another Datacenter. Is it true?
If it is true , How Can I Config a cluster between 2 separaqte datacenter , that we dont have any privatre network between them ?
Thank you .
This is our suggesstion Plan ?

I hope I understand your question correctly, if you would like to setup alwayson AG across 2 DC (Datacenter), you will have to make sure they can be connected. It can be on 2 different subnet (when you use Windows 2012 onwards), and SQL 2012 supports it.

You can still have a alwayson replica (sync) to be on the same site (to provide HA), and another async node to provide for DR (on another DC). This might can cater your network latency between the 2 sites. Since WSFC is required for alwayson, I believe you will need to be on the same domain, if you do not have any connectivity between the 2 sites, I don't think you will be able to setup alwayson on them.

Thank you Dennis
We have connectivity between 2 DC. But It is Public Connectivity (means Internet) . For instance we can connect to second DC and see and manage some servers now. But This connectivity is based on internet .
we dont have any private connectivity between main DC And second DC. For example we dont have Private fiber cable or Wireless connection between 2 DCs.
in this situation, Can we have cluster between them for Alwayson. Or not ?

as long as both servers are connected in the same domain, I believe so. However, given you are on public internet, your best bet will be using async mode. Given its pubic network, you might need more IP to get WSFC working, I haven't try it myself so I'm just guessing here. Given you have enough IPs, and all belongs to same domain, with connectivity, I think you should be able to.

If you can get WSFC going, i think you should be able to get alwayson going.