Understanding Red-Black Trees: Properties, Height, and Rotations

1. CMPT 225 Red–black trees

2. Red-black Tree Structure • A red-black tree is a BST! • Each node in a red-black tree has an extra color field which is • red or • black • In addition false nodes are added so that every (real) node has two children • These are pretend nodes, they don’t have to have space allocated to them • These nodes are colored black • We do not count them when measuring a height of nodes • Nodes have an extra reference to their parent

3. Red-black Tree Properties • 1 – Every node is either red or black • 2 – Every leaf is black • This refers to the pretend leaves • In implementation terms, every null child of a node considered to be a black leaf • 3 – If a node is red both its children must be black • 4 – Every path from a node to its descendent leaves contains the same number of black nodes • 5 – The root is black (mainly for convenience)

4. 87 65 47 32 93 82 71 Example of RB tree

5. RB Trees: Data structure publicclass TreeNode<T> { publicenum Color {red, black} private T item; private TreeNode<T> leftChild; private TreeNode<T> rightChild; private TreeNode<T> parent; private Color color; // constructors, accessors, mutators... }

6. Red-black Tree Height • The black height of a node, bh(v), is the number of black nodes on any path from v to a leaf (not counting the pretend node). • Remember that every path from a node to a leaf contains the same number of black nodes • The height of a node, h(v), is the number of nodes on the longest path from v to a leaf (not counting the pretend node). • Let bh be the black height of the tree (root) and h the height of the tree (root)

7. Analysis of Red-black Tree Height Assume that a tree has n internal nodes • An internal node is a non-leaf node (a node containing a value), remember that the leaf nodes are just pretend nodes • Claim: The tree will have h 2*log(n+1) • A tree has at least 2bh – 1 internal (real) nodes, i.e., n  2bh – 1 • Proof by MI • bh  h / 2 (from property 3 of a RB-tree) • Hence,n  2h/2 – 1 • log(n + 1)  h / 2 (+1 and take log2 of both sides) • h  2*log(n + 1) (multiply by 2)

8. Rotations • An item must be inserted into a red-black tree at the correct position • The shape of any particular tree is determined by: • the values of the items inserted into the tree • the order in which those values are inserted • A tree’s shape can be altered by rotation while still preserving the bst property

9. x z z x A C B C B A Left Rotation Left rotate(x)

10. x z z x A C B C B A Right Rotation Right rotate(z)

11. x x z z 47 81 32 13 40 37 44 C B A C B A Left Rotation Example Perform a left rotation of the node with the value 32, which we will call x - Make a reference to the right child of x x z

12. x x z z 37 40 44 32 81 47 13 C B A B C A Left Rotation Example Perform a left rotation of the node with the value 32, which we will call x - Make a reference to the right child of x x - Make z’s left child x’s right child - Detach z’s left child z

13. x z x z 81 32 13 40 37 47 44 B C A B C A Left Rotation Example Perform a left rotation of the node with the value 32, which we will call x - Make a reference to the right child of x x - Make z’s left child x’s right child - Detach z’s left child - Make x the left child of z z - Make z the child of x’s parent

14. x z x z 81 32 13 40 37 47 44 B C A B C A Left Rotation Example Perform a left rotation of the node with the value 32, which we will call x - Make a reference to the right child of x x - Make z’s left child x’s right child - Detach z’s left child - Make x the left child of z z - Make z the child of x’s parent

15. x x z z 47 81 40 32 44 13 37 C B A B A C Left Rotation Example After Performing a left rotation of the node with the value 32 z x

16. Left Rotation: Algorithm TreeNode<T> leftRotate(TreeNode<T> x) // returns a new root of the subtree originally rooted in x // (that is: returns a reference to z) // Pre: rigth child of x is a proper node (with value) { TreeNode<T> z = x.getRight(); x.setRight(z.getLeft()); // Set parent reference if (z.getLeft() != null) z.getLeft().setParent(x); z.setLeft(x); //move x down z.setParent(x.getParent()); // Set parent reference of x if (x.getParent() != null) //x is not the root if (x == x.getParent().getLeft()) //left child x.getParent().setLeft(z); else x.getParent().setRight(z); x.setParent(z); return z; }

17. z z x x 37 47 81 32 7 40 13 29 B B A C A C Right Rotation Example Perform a right rotation of the node with the value 47, which we will call z - Make a reference to the left child of z x

18. z x z x 47 81 32 40 37 7 29 13 A B B C A C Right Rotation Example Perform a right rotation of the node with the value 47, which we will call z - Make a reference to the left child of z - Make x’s right child z’s left child x • Detach x’s right child

19. x z z x 29 7 37 13 32 81 47 40 A C B B A C Right Rotation Example Perform a right rotation of the node with the value 47, which we will call z - Make a reference to the left child of z - Make x’s right child z’s left child x - Detach x’s right child - Make z the left child of x

20. z z x x 81 37 29 7 13 40 47 32 B C B A A C Right Rotation Example Perform a right rotation of the node with the value 47, which we will call z - Make a reference to the left child of z - Make x’s right child z’s left child x - Detach x’s right child - Make z the left child of x - In this case, make x the new root

21. RB Tree: Operations • Search: the same as Search for BST (cannot spoil RB properties) • Insertion: first insert with BST insertion algorithm, then fix RB properties • Deletion: first delete with BST deletion algorithm, then fix RB properties To fix use: • rotate operations • recoloring of nodes -> neither of them will spoil the BST property Demonstration of Red/Black Tree operations: http://www.ece.uc.edu/~franco/C321/html/RedBlack/

22. Red-black Tree Insertion • First perform the standard BST insertion • Color the inserted node x with color red • The only r-b property that can be violated is #3, that both a red node’s children are black • If x’s parent p is black, we are done • Assume it’s red: • If x’s uncle is red, recolor x’s parent, its parent and uncle, and continue recursively • If x’s uncle is black, first make sure x is left child (left_rotate in p if necessary); second right_rotate in parent of p

23. x’s uncle is red

24. x’s uncle is black left_rotate on p • Assuming that p is left child of its parent, in the other case, everything is symmetric p x right_rotate on p^2[x]