1. CS 473Lecture X1 CS473-Algorithms Lecture RED-BLACK TREES (RBT)

2. CS 473Lecture X2 Overview Previous lecture showed that a BST of height h can implement any of the dynamic set operations in O(h) time. Thus, set operations are fast if the height of BST is small. But, if h is large, its performance is no better than linked list.

3. CS 473Lecture X3 Overview Red-Black trees are one of many search tree schemes that are balanced to guarantee h= O(lgn) First, we will show that h= O(lgn) in RBTs Then, show how to manipulate & maintain RBTs during dynamic set operations.

4. CS 473Lecture X4 Properties of Red-Black Trees Add color field (1 bit) to each node of a BST. Consider “NIL” child fields as pointer to external nodes (leafs) i.e. Leaf nodes do not have keys Consider normal key-bearing nodes as internal nodes. Hence, each internal node has exactly 2 children. Thus, RBTs are full binart trees (FBTs) FBT: Each node is either a leaf or has degree exactly 2. i.e. There are no degree-1 nodes.

5. CS 473Lecture X5 Red-Black Properties 1.Every node is either red or black 2.Every leaf (NIL) is black 3.If a node is red, then both of its children are black. i.e. There can’t be 2 consecutive reds on a simple path. 4.Every simpe path from a node to a descendant leaf contains the same number of black nodes. 5.( Additional propery) The root is black. 6.Simplifies explanation of insertion which assumes and guarantees that the root is black.

6. CS 473Lecture X6 A Sample Red Black Tree bh = 2 h = 4 bh = 2 h = 3 bh = 1 h = 2 bh = 1 h = 1 bh = 0 h = 0 bh = 1 h = 1 bh = 2 h = 4

7. CS 473Lecture X7 Properties of Red-Black Trees Black-Height : bh (x) # of blacks on the simple paths to all descendant leaves not counting x. By the property 4, the black-height is well defined. Red-Black properties ensure that No simpe path from a node to a descendant leaf is more that twice as long as any other. So, red-black trees are apptoximately balanced.

8. CS 473Lecture X8 Properties of Red-Black Trees Lemma 1 : A height h node has black height ≥ h/2 –Height is length of the longest path to a leaf. –By property 3, # of reds on this path ≤ h/2 –Hence, # of blacks ≥ h/2 Lemma 2 : The subtree T x rooted at any node x contains at least |T x | ≥ 2 bh(x) – 1 internal nodes.

9. CS 473Lecture X9 Properties of Red-Black Trees Basis: h(x)=0 x must be a leaf (NIL) bh(x)=0 T x contains 0 internal nodes = 2 bh(x) -1=2-1= 0 Inductive Step: h(x) > 0 x is internal node with 2 children l[x] : left[x] r[x] : right[x] h(x) bh(x) r[x] l [x] T r[x] T l[x] x

10. CS 473Lecture X10 Depending on l[x] and r[x] being red or black –bh(l[x]) = bh(x) or bh(x) -1, respectively, –bh(r[x]) = bh(x) or bh(x) -1, respectively. Since h(l[x]), h(r[x]) < h(x) we can apply inductive hypothesis. |T x | = |T l | + |T r | +1 ≥ (2 bh(x)-1 – 1) + (2 bh(x)-1 – 1) + 1 = 2 bh(x)-1 – 1 Q.E.D. Properties of Red-Black Trees

11. CS 473Lecture X11 Theorem : A red-black tree with n internal nodes has height at most 2lg(n+1) Due to lemma 2 |T root | = n ≥ 2 bh(root) – 1 Due to lemma 1 bh(root) ≥ h(root) / 2 = H / 2 n ≥ 2 H/2 -1 (n+1) ≥ 2 H/2 H ≤ 2 lg(n+1) Properties of Red-Black Trees

12. CS 473Lecture X12 Corollary: Dynamic set operations MIN, MAX, SEARCH, SUCCESSOR, PREDECESSOR take O(lgn) time. Fact: Let l min (x) & l max (x) denote the lengths of the shortest & longest paths from a node x to its leafs, respectively. Then, l max (x) ≤ l min (x) for any node x. Proof: l max (x) = h(x) ≤ 2bh(x) ≤ l min (x) Properties of Red-Black Trees Lemma 1

13. CS 473Lecture X13 The Problem of Insertion into Red-Black Trees Draw a R-B tree Color choices starting at 12 Forced choices : 12 → R ; 5 and 9 → B 7 5 12 9

14. CS 473Lecture X14 The Problem of Insertion into Red-Black Trees Insert 8 (Left child of 9) No problem, Just color it red 7 5 12 9 8 ? → R

15. CS 473Lecture X15 The Problem of Insertion into Red-Black Trees Insert 11 (Left child of 12) 11 can’t be red (Violate property 3) 11 can’t be black (Violate property 4) 7 5 12 9 8 11 ?

16. CS 473Lecture X16 The Problem of Insertion into Red-Black Trees Can fix up by recoloring the tree New 11 to Red Change 9 to Red Change 8 and 12 to Black 7 5 12 9 8 R → B B → R R → B 11 ?→R

17. CS 473Lecture X17 The Problem of Insertion into Red-Black Trees Insert 10 (Left child of 11) Recoloring is not enough, why? -Because of tree imbalance -Can’t satisfy property 4 (Equal # of blacks on paths) without violating property 3 (No consecutive reds on paths) Must change tree structure -Goal : Restructure in O(lgn) time. 7 5 12 9 8 11 10 ?