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Foundation of Computing SystemsPowerPoint Presentation

Foundation of Computing Systems

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### … Continued

IT 60101: Lecture #11

Deletion in a Red-Black Tree

- Deletion of a node from a red-black tree follows two major steps
- Delete the node using the usual deletion operation as in a binary search tree.
- This deletion first finds the location of the node to be deleted and then it is replaced by its inorder successor.
- If the inorder successor has its child, then the child is moved up to the place of the inorder successor.
2. The above deletion operation may disturb the properties of a red-black tree. To restore the red-black tree, we require a fix-up operation.

IT 60101: Lecture #11

Deletion in a Red-Black Tree

- Suppose the node that has to be deleted is x and its inorder successor is y.
- The following situation and steps to be followed accordingly as discussed below.
Case 1:y is NULL, that is, node x does not have any inorder successor.

Case 2: y is RED and y has no children as internal nodes

Case 3: y is BLACK and it has no children as an internal node

Case 4:y is BLACK and it has an internal RED node

IT 60101: Lecture #11

Deletion Operation: Case 1

- Case 1:y is NULL, that is, node x does not have any inorder successor.

IT 60101: Lecture #11

Deletion Operation: Case 1

- Case 1:y is NULL, that is, node x does not have any inorder successor.

IT 60101: Lecture #11

Deletion Operation: Case 1

- Case 1:y is NULL, that is, node x does not have any inorder successor.

IT 60101: Lecture #11

Deletion Operation: Case 2

- Case 2: y is RED and y has no children as internal nodes

IT 60101: Lecture #11

Deletion Operation: Case 2

- Case 2: y is RED and y has no children as internal nodes

IT 60101: Lecture #11

Deletion Operation: Case 3

- Case 3: y is BLACK and it has no children as an internal node

IT 60101: Lecture #11

Deletion: Fix-up Operation

- Suppose, x is the node to be deleted. y is its inorder successor and z be its another child other than the leaf child.
- The following three problems may arise while x is deleted from the red-black tree.
- y is red and it becomes the new root. This violates the “root property”.
- Both z and the parent of y (now after deletion of y the parent of y becomes the parent of z) are red. This violates the “internal property” and leads to a “double-red problem”.
- y’s removal causes any path that contained one fewer black node than the path before the removal of y. This violates the “black-depth property”.

IT 60101: Lecture #11

Deletion: Fix-up Operation

- Assume that z is the left sub-tree with respect to the parent y. The following four cases can occur while fixing up the above-mentioned problem.
- Case 1: z’s sibling w is RED
- Case 2: z’s sibling w is BLACK and both the children of w are BLACK
- Case 3:z’s sibling w is BLACK and w’s left child is RED and w’s right child is BLACK.
- Case 4: z’s sibling w is BLACK and w’s RIGHT child is RED.

IT 60101: Lecture #11

Fix-up Operation: Case 1

- Case 1: z’s sibling w is RED
- Steps:
- Change the colour of w to BLACK
2. Switch (flip) the colour of parent of z.

3. Perform a left rotation of w

4. Sibling of z becomes a new sibling

- Change the colour of w to BLACK
- This may lead to any one of the following problems of fix-up Case 2, Case 3 and Case 4 (when the node w is black).
- These three cases are distinguished by the colours of children of w.

IT 60101: Lecture #11

Fix-up Operation: Case 1

IT 60101: Lecture #11

Fix-up Operation: Case 2

- Case 2: z’s sibling w is BLACK and both the children of w are BLACK
- Steps:
- Make the colour w of RED
- Add an extra black to the parent of z (which was earlier either red or black).
- Repeat the fix-up procedure but now changing the parent of z becomes a new z.

IT 60101: Lecture #11

Fix-up Operation: Case 2

IT 60101: Lecture #11

Fix-up Operation: Case 3

- Case 3:z’s sibling w is BLACK and w’s left child is RED and w’s right child is BLACK.
- Steps:
- Exchange the colours of w and its left child.
- Perform a right rotation of w’s left child. This rotation results in a new sibling w of z which is a black node with a red child.

IT 60101: Lecture #11

Fix-up Operation: Case 3

IT 60101: Lecture #11

Fix-up Operation: Case 4

- Case 4: z’s sibling w is BLACK and w’s RIGHT child is RED.
- Steps:
- Exchange the colour of w with the colour of parent of z.
- Perform a left-rotation of parent of z (or w)
- Repeat the fix-up procedure until z reaches the root or z becomes BLACK.

IT 60101: Lecture #11

Fix-up Operation: Case 4

IT 60101: Lecture #11

- For detail implementation of deletion operation on red-black binary trees see the book
- Classic Data Structures
- Chapter 7
- PHI, 2nd Edn., 17th Reprint

IT 60101: Lecture #11

HB (AVL) Tree vs. RB Tree binary trees see the book Specialization: Generalization:

- Height
- A red-black tree with n internal nodes has height at most 2log2(n+1)
- An AVL tree with n internal nodes has height at most 1.44log2n (see Lemma 7.11).

- An AVL tree is a kind of a red-black tree (in other words, all AVL trees satisfy the properties of a red-black tree but all red-black trees are not AVL trees).

- A red-black tree is a kind of B-tree. This is why a red-black tree is also alternatively termed Symmetric B-tree. (see Section 7.5.9: How a red-black tree is a kind of B-tree is discussed in the next section).

IT 60101: Lecture #11

Application of RB Tree binary trees see the book

- A red-black tree is used to improve the speed of searching in a B-tree
- Restructuring with RB tree

IT 60101: Lecture #11

Restructuring with RB Tree binary trees see the book

IT 60101: Lecture #11

Restructuring with RB Tree binary trees see the book

IT 60101: Lecture #11

Restructuring with RB Tree binary trees see the book

IT 60101: Lecture #11

Restructuring with RB Tree binary trees see the book

IT 60101: Lecture #11

Restructuring with RB Tree binary trees see the book

IT 60101: Lecture #11

Restructuring with RB Tree binary trees see the book

IT 60101: Lecture #11

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