A simple find should be enough:

find /path/to/dir -type d -empty

For example:

$ tree foo

foo

├── 1

│   ├── 1

│   ├── 2

│   └── 3

├── 2

│   ├── 1

│   ├── 2

│   └── 3

└── 3

    ├── 1

    ├── 2

    └── 3



12 directories, 0 files

$ find foo -type d -empty

foo/2/2

foo/2/3

foo/2/1

foo/3/2

foo/3/3

foo/3/1

foo/1/2

foo/1/3

foo/1/1

$ touch foo/1/1/a foo/2/1/a foo/3/1/a 

$ find foo -type d -empty            

foo/2/2

foo/2/3

foo/3/2

foo/3/3

foo/1/2

foo/1/3

If these directories can contain files, then this would be better, but expensive:

find foo -type d -exec sh -c 'find "$0" -mindepth 1 -type d -printf z | grep -q z || printf "%s

" "$0"' {} ;

This Stack Overflow post has a rather neat solution:

find /path/to/dir -type d -links 2

Since a directory without subdirectories only has two hard links, one two its parent and one to itself.

The question, phrased differently, has been asked before on Stack Overflow, Super User and Unix & Linux:

• Use GNU find to show only the leaf directories


• List all leaf subdirectories in linux


• Using “find” to list only directories with no more childs


• find all end subdirectories in a tree