if n has 2 children, let x be node in n's left sub tree with the smallest key. remove x (since x has no left child and is easily rmvd). and now replace n's key with x's key.

if n has 2 children, let x be node in n's right sub tree with the largest key. remove x (since x has no left child and is easily rmvd). and now replace n's key with x's key.

if n has 2 children, let x be node in n's right sub tree with the smallest key. remove x (since x has no left child and is easily rmvd). and now replace n's key with x's key.

if n has 2 children, let x be node in n's left sub tree with the largest key. remove x (since x has no left child and is easily rmvd). and now replace n's key with x's key.

Answer :(C)

if n has 2 children, let x be node in n's right sub tree with the smallest key. remove x (since x has no left child and is easily rmvd). and now replace n's key with x's key.

Description :

if n has 2 children, let x be node in n's right sub tree with the smallest key. remove x (since x has no left child and is easily rmvd). and now replace n's key with x's key. There are other ways also to do it, but this is most efficient way to do it with out disturbing overall tree weight and distributing all the nodes properly.