不用find，怎样递归地给目录设置700，给文件设置600权限？

https://stackoverflow.com/questions/36553701/how-to-set-permissions-recursively-700-for-folders-and-600-for-files-without-u

I'm trying to figure out a way to set permissions recursively 700 for dirs and subdirs on a specific path and 600 for files. I would use these commands:

find /path -type d -print0 | xargs -0 chmod 700

find /path -type f -print0 | xargs -0 chmod 600

But the user does not have permission to run the "find" command. As a workaround I tried to make a script that contains the above commands from the root user with setuid sticky bit set so it will run with root privileges (like passwd or sudo commands that normal users run with root privileges):

chmod 4755 script.sh

but i cannot execute the script from the limited user account, it still says that I don't have permission to run the find command.

Does anyone have any idea how i can accomplish this without having to use the "find" command?

Edit: OS: Centos 6.5

Apparently this is very easy to implement. There are 2 ways: using chmod only, or setting ACL (access control list) on the desired path:

Using chmod i would run:

chmod -R u=rwX,g=,o= /path

for the user owner i'm giving capital "X", so it does apply only to directories and not files.

Using ACL:

setfacl -Rm u::rwX,g::0,o::0 /path

setfacl -Rm d:u::rwX,g::0,o::0 /path

again using capital "X" so it applies only to directories and not files. The first command applies the ACL, the second one makes it default policy so newly created files will inherit the desired permissions.