[BST] Deletion procedure
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by Metz
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Published on 2010-06-07T14:46:23Z
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2010/06/07
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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.
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