[BST] Deletion procedure

Posted by Metz on Stack Overflow See other posts from Stack Overflow or by Metz
Published on 2010-06-07T14:46:23Z Indexed on 2010/06/07 17:42 UTC
Read the original article Hit count: 246

Hi all.

Consider the deletion procedure on a BST, when the node to delete has two children. Let's say i always replace it with the node holding the minimum key in its right subtree.

The question is: is this procedure commutative? That is, deleting x and then y has the same result than deleting first y and then x?

I think the answer is no, but i can't find a counterexample, nor figure out any valid reasoning.

Thank you.

EDIT:

Maybe i've got to be clearer.

Consider the transplant(node x, node y) procedure: it replace x with y (and its subtree). So, if i want to delete a node (say x) which has two children i replace it with the node holding the minimum key in its right subtree:

y = minimum(x.right)
transplant(y, y.right) // extracts the minimum (it doesn't have left child)
y.right = x.right
y.left = x.left
transplant(x,y)

The question was how to prove the procedure above is not commutative.

© Stack Overflow or respective owner

Related posts about data-structures

Related posts about binary-tree