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#### Published on 24 Mar 2014 | over 4 years ago

See the rules of Red Black Tree's and each violation case and their respective solution below and check out my other video about this: youtu.be/axa2g5oOzCE . Also, see below for descriptions of inserting and deleting nodes from a Red Black Tree.

Rules of Red Black Trees (not in any particular order):
1. Every node is either RED or BLACK
2. The root node is BLACK
3. Every leaf node (null) is BLACK
4. Every RED node has two BLACK child nodes
5. Every path from node x to a leaf node has the same number of BLACK nodes (BLACK height)

VIOLATIONS and SOLUTIONS:
Case 1:
1. Uncle is Red

Solution for Case 1:
1. Swap colors of Parent, Uncle, and Grandparent
2. Grandparent becomes the new node to check for violations

Case 2:
1. Uncle is Black
2a. Node is a Left Child of a Right Child
2b. Node is a Right Child of a Left Child

Solution for Case 2:
1a. Right rotate around Parent for 2a = Case 3
1b. Left rotate around Parent for 2b = Case 3
2. Parent becomes the new node to check for violations

Case 3:
1. Uncle is Black
2a. Node is a Right Child of a Right Child
2b. Node is a Left Child of a Left Child

Solution for Case 3:
1a. Left rotate around Grandparent for 2a
1b. Right rotate around Grandparent for 2b
2. Swap colors of Parent and Grandparent
3. Grandparent becomes the new node to check for violations

INSERTION
To insert a node, you begin at the root of the tree, and follow the rules of a Binary Search Tree, to insert the new node. The only difference is that you have to give the node a starting color (RED) and then after the insertion, check for violations (as described above).

DELETION
To delete a node, you follow the rules of deletion for Binary Search Trees. There are three scenarios:

Assuming we want to delete node x:
1. If node x has no children, then just delete node x.
2. If node x has one child, then replace node x with it's child, and delete node x.
3. If node x has two children, then replace node x with either its predecessor or successor, and delete node x.

For scenario 2 or 3 above, you would then check for violations at the node that replaced node x.
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