Display only a process subtree

Question

I would like to show a listing of a single process and its current children. So, given the following process tree:

Imagine the following process listing:

  PID TTY      STAT   TIME COMMAND
    2 ?        S      0:00 [kthreadd]
    3 ?        S      0:06  \_ [ksoftirqd/0]  
  ...snip...
 1292 ?        Ss     0:06 /usr/sbin/gpm -m /dev/input/mice -t exps2
 1426 ?        Ss     0:00 /usr/lib/postfix/master
 9785 ?        S      0:00  \_ qmgr -l -t fifo -u
12301 ?        S      0:00  \_ pickup -l -t fifo -u -c
 1545 ?        Ss     0:05 /usr/sbin/apache2 -k start
 1570 ?        S      0:00  \_ /usr/sbin/apache2 -k start  
  ...snip...

I would like to instead just show process 1426 and its children. Like this:

  PID TTY      STAT   TIME COMMAND
 1426 ?        Ss     0:00 /usr/lib/postfix/master
 9785 ?        S      0:00  \_ qmgr -l -t fifo -u
12301 ?        S      0:00  \_ pickup -l -t fifo -u -c

Is there a simple way to do this?

Answer 1

You can use pstree to do this and get a nicely formatted output too

pstree -p 22221
mysqld_safe(22221)─┬─logger(22334)
                   └─mysqld(22332)─┬─{mysqld}(22335)
                                   ├─{mysqld}(22336)
                                   ├─{mysqld}(22337)
                                   ├─{mysqld}(22338)
                                   ├─{mysqld}(22340)
                                   ├─{mysqld}(22341)
                                   ├─{mysqld}(22342)
                                   ├─{mysqld}(22343)
                                   ├─{mysqld}(22346)
                                   └─{mysqld}(22394)

Answer 2

$ ps -p 1426 --ppid 1426 --forest

or:

$ ps -eo pid,ppid,tty,stat,time,command --forest | awk '$1 == 1426 || $2 == 1426'

to display the details command.