Red Black Trees Colored Nodes Definition Binary search tree. Each node is colored red or black. Root and all external nodes are black. No root-to-external-node path has two consecutive red nodes. All root-to-external-node paths have the same number of black nodes Every red node has a black parent.

Red Black Trees Colored Edges Definition Binary search tree. Child pointers are colored red or black. Pointer to an external node is black. No root to external node path has two consecutive red pointers. Every root to external node path has the same number of black pointers.

Example Red-Black Tree 10 7 8 1 5 30 40 20 25 35 45 60 3 Height = 5

Properties The height of a red black tree that has n (internal) nodes is between log2(n+1) and 2log2(n+1).

Properties Start with a red black tree whose height is h; collapse all red nodes into their parent black nodes to get a tree whose node-degrees are between 2 and 4, height is >= h/2, and all external nodes are at the same level.

Properties 10 7 8 1 5 30 40 20 25 35 45 60 3 Collapsed height = 3.

Properties Let h’>= h/2 be the height of the collapsed tree. In worst-case, all internal nodes of collapsed tree have degree 2. Number of internal nodes in collapsed tree >= 2h’-1. So, n >= 2h’-1 So, h <= 2 log2 (n + 1)

Properties At most 1 rotation and O(log n) color flips per insert/delete. Priority search trees. Two keys per element. Search tree on one key, priority queue on other. Color flip doesn’t disturb priority queue property. Rotation disturbs priority queue property. O(log n) fix time per rotation => O(log2n) overall time. Color flip to be defined later. Binary search trees in which each node contains a binary search tree are used, for example, in IP router tables. Each rotation of the outer binary search tree requires changes in the binary search trees of the involved nodes resulting in an O(log n) cost per rotation.

Properties O(1) amortized complexity to restructure following an insert/delete. C++ STL implementation java.util.TreeMap => red black tree

Insert New pair is placed in a new node, which is inserted into the red-black tree. New node color options. Black node => one root-to-external-node path has an extra black node (black pointer). Hard to remedy. Red node => one root-to-external-node path may have two consecutive red nodes (pointers). May be remedied by color flips and/or a rotation.

Classification Of 2 Red Nodes/Pointers b c d gp pp p LLb XYz X => relationship between gp and pp. pp left child of gp => X = L. Y => relationship between pp and p. p right child of pp => Y = R. z = b (black) if d = null or a black node. z = r (red) if d is a red node. Initially, p is newly inserted node. Pp and gp may be null. In each round, p, pp, and gp are moved up by 2 levels. If p is a red root, make it black. If p or pp black done! If pp is a red root, make it black. O.w., gp exists

XYr Color flip. Move p, pp, and gp up two levels. b c d gp pp p a b c d gp pp p Move p, pp, and gp up two levels. Continue rebalancing if necessary.

LLb Rotate. Done! Same as LL rotation of AVL tree. a b c d a b c d gp y x a b z c d a b c d gp pp p x y z Done! Same as LL rotation of AVL tree.

LRb Rotate. Done! Same as LR rotation of AVL tree. y x a b z c d b c a d gp pp p y x z Done! Same as LR rotation of AVL tree. RRb and RLb are symmetric.

Delete Delete as for unbalanced binary search tree. If red node deleted, no rebalancing needed. If black node deleted, a subtree becomes one black pointer (node) deficient.

Delete A Black Leaf 10 7 8 1 5 30 40 20 25 35 45 60 3 Delete 8.

Delete A Black Leaf y is root of deficient subtree. py is parent of y. 10 7 1 5 30 40 20 25 35 45 60 3 py y If py doesn’t exist, the whole tree is one black node deficient. This is OK. y is root of deficient subtree. py is parent of y.

Delete A Black Degree 1 Node 10 7 8 1 5 30 40 20 25 35 45 60 3 py y Delete 45. y is root of deficient subtree.

Delete A Black Degree 2 Node 10 7 8 1 5 30 40 20 25 35 45 60 3 Not possible, degree 2 nodes are never deleted.

Rebalancing Strategy If y is a red node, make it black. py y 10 7 8 1 5 30 40 20 25 35 45 60 3 y py

Rebalancing Strategy Now, no subtree is deficient. Done! py y 60 10 7 8 1 5 30 40 20 25 35 45 3 y py

Rebalancing Strategy y is a black root (there is no py). Entire tree is deficient. Done! 60 10 7 8 1 5 30 40 20 25 35 45 3 y

Rebalancing Strategy y is black but not the root (there is a py). Xcn v Xcn y is right child of py => X = R. Pointer to v is black => c = b. v has 1 red child => n = 1. y represents both a subtree and its root. This includes the case when y is null. Broken lines indicate pointers whose colors are unspecified. v must exist because y is one black node/pointer deficient.

Rb0 (case 1) Color change. Now, py is root of deficient subtree. v a b y py v Color change. Now, py is root of deficient subtree. Continue!

Rb0 (case 2) Color change. Deficiency eliminated. Done! y a b py v a b

Rb1 (case 1) LL rotation. Deficiency eliminated. Done! a b y v py a b

Rb1 (case 2) LR rotation. Deficiency eliminated. Done! c y w py a b v

Rb2 c y w py a b v a y py v b c w LR rotation. Deficiency eliminated. Done!

Rr(n) n = # of red children of v’s right child w. a y py v b c w Rr(2)

Rr(0) a b y v py a b y py v LL rotation. Done!

Rr(1) (case 1) LR rotation. Deficiency eliminated. Done! y c w py a v b a y py v b w c LR rotation. Deficiency eliminated. Done!

Rr(1) (case 2) y d x py a v b c w a y py v b w c d x Rotation. Deficiency eliminated. Done!

Rr(2) Rotation. Deficiency eliminated. Done! d y x py a v b c w a y py

B-Trees Large degree B-trees used to represent very large dictionaries that reside on disk. Smaller degree B-trees used for internal-memory dictionaries to overcome cache-miss penalties. May also be used internally to reduce cache misses and hence improve performance.

AVL Trees n = 230 = 109 (approx). 30 <= height <= 43. When the AVL tree resides on a disk, up to 43 disk access are made for a search. This takes up to (approx) 4 seconds. Not acceptable. 43 cache misses possible when searching memory resident AVL trees

Red-Black Trees n = 230 = 109 (approx). 30 <= height <= 60. When the red-black tree resides on a disk, up to 60 disk access are made for a search. This takes up to (approx) 6 seconds. Not acceptable.

m-way Search Trees Each node has up to m – 1 pairs and m children. m = 2 => binary search tree.

4-Way Search Tree 10 30 35 k > 35 k < 10 10 < k < 30 10 30 35 k > 35 k < 10 10 < k < 30 30 < k < 35

Maximum # Of Pairs Happens when all internal nodes are m-nodes. Full degree m tree. # of nodes = 1 + m + m2 + m3 + … + mh-1 = (mh – 1)/(m – 1). Each node has m – 1 pairs. So, # of pairs = mh – 1.

Capacity Of m-Way Search Tree Natural extension of 2-3 and 2-3-4 trees would leave worst case at m=2.

Definition Of B-Tree Definition assumes external nodes (extended m-way search tree). B-tree of order m. m-way search tree. Not empty => root has at least 2 children. Remaining internal nodes (if any) have at least ceil(m/2) children. External (or failure) nodes on same level. Alternative definition has pairs only in leaves; remaining nodes have keys only. Leaf-pushed B-tree or B+-tree.

2-3 And 2-3-4 Trees B-tree of order m. 2-3 tree is B-tree of order 3. m-way search tree. Not empty => root has at least 2 children. Remaining internal nodes (if any) have at least ceil(m/2) children. External (or failure) nodes on same level. 2-3 tree is B-tree of order 3. 2-3-4 tree is B-tree of order 4.

B-Trees Of Order 5 And 2 B-tree of order m. m-way search tree. Not empty => root has at least 2 children. Remaining internal nodes (if any) have at least ceil(m/2) children. External (or failure) nodes on same level. Not possible for an internal node in an extended tree to have only 1 child. B-tree of order 5 is 3-4-5 tree (root may be 2-node though). B-tree of order 2 is full binary tree.

Minimum # Of Pairs n = # of pairs. # of external nodes = n + 1. Height = h => external nodes on level h + 1. level # of nodes 1 1 >= 2 2 >= 2*ceil(m/2) 3 >= 2*ceil(m/2)h-1 h + 1 n + 1 >= 2*ceil(m/2)h-1, h >= 1

Minimum # Of Pairs n + 1 >= 2*ceil(m/2)h-1, h >= 1 m = 200. height # of pairs >= 199 2 >= 19,999 3 >= 2 * 106 – 1 4 >= 2 * 108 – 1 5 h <= log ceil(m/2) [(n+1)/2] + 1

Choice Of m Worst-case search time. search time m (time to fetch a node + time to search node) * height (a + b*m + c * log2m) * h where a, b and c are constants. m search time 50 400 Typically pick m so as to fill a block.

Insert 8 4 15 20 1 3 5 6 9 30 40 16 17 2-3 Tree. Insert 10, no problem. Insert 18? Insertion into a full leaf triggers bottom-up node splitting pass.

Split An Overfull Node m a0 p1 a1 p2 a2 … pm am ai is a pointer to a subtree. pi is a dictionary pair. ceil(m/2)-1 a0 p1 a1 p2 a2 … pceil(m/2)-1 aceil(m/2)-1 Node Splitting m-ceil(m/2) aceil(m/2) pceil(m/2)+1 aceil(m/2)+1 … pm am pceil(m/2) plus pointer to new node is inserted in parent.

Insert 8 4 15 20 1 3 5 6 9 30 40 16 17 Insert a pair with key = 2. 15 20 1 3 5 6 9 30 40 16 17 Insert a pair with key = 2. New pair goes into a 3-node.