Red-Black Trees—Again rank(x) = # black pointers on path from x to an external node. Same as #black nodes (excluding x) from x to an external node. rank(external.

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Red-Black Trees—Again rank(x) = # black pointers on path from x to an external node. Same as #black nodes (excluding x) from x to an external node. rank(external node) = 0.

An Example

Properties Of rank(x) rank(x) = 0 for x an external node. rank(x) = 1 for x parent of external node. p(x) exists => rank(x) <= rank(p(x)) <= rank(x) + 1. g(x) exists => rank(x) < rank(g(x)).

Red-Black Tree  A binary search tree is a red-black tree iff integer ranks can be assigned to its nodes so as to satisfy the stated 4 properties of rank.

Relationship Between rank() And Color (p(x),x) is a red pointer iff rank(x) = rank(p(x)). (p(x),x) is a black pointer iff rank(x) = rank(p(x)) – 1. Red node iff pointer from parent is red. Root is black. Other nodes are black iff pointer from parent is black. Given rank(root) and node/pointer colors, remaining ranks may be computed on way down.

rank(root) Height <= 2 * rank(root). No external nodes at levels 1, 2, …, rank(root).  So, #nodes >=  1 <= i <= rank(root) 2 i -1 = 2 rank(root) – 1.  So, rank(root) <= log 2 (n+1). So, height(root) <= 2log 2 (n+1).

Join(S,m,B) Input  Dictionary S of pairs with small keys.  Dictionary B of pairs with big keys.  An additional pair m.  All keys in S are smaller than m.key.  All keys in B are bigger than m.key. Output  A dictionary that contains all pairs in S and B plus the pair m.  Dictionaries S and B may be destroyed.

Join Binary Search Trees O(1) time. S m B

Join Red-black Trees When rank(S) = rank(B), use binary search tree method. S m B rank(root) = rank(S) + 1 = rank(B) + 1.

rank(S) > rank(B) Follow right child pointers from root of S to first node x whose rank equals rank(B). ab x p(x) S ab x m B

rank(S) > rank(B) If there are now 2 consecutive red pointers/nodes, perform bottom-up rebalancing beginning at m. O(rank(S) – rank(B)). ab x p(x) S ab x m B

rank(S) < rank(B) Follow left child pointers from root of B to first node x whose rank equals rank(B). Similar to case when rank(S) > rank(B).

Split(k) Inverse of join. Obtain  S … dictionary of pairs with key < k.  B … dictionary of pairs with key > k.  m … pair with key = k (if present).

Split A Binary Search Tree b A a B c C d D e E f m g SB

b A a B c C d D e E f m g BS

b A a B c C d D e E f m g BS

b A a B c C d D e E f m g BS

b A a B c C d D e E f m g BS

b A a B c C d D e E f m g BS

b A a B c C d D e E f m g BS

Split A Red-Black Tree Previous strategy does not split a red-black tree into two red-black trees. Must do a search for m followed by a traceback to the root. During the traceback use the join operation to construct S and B.

Split A Red-Black Tree b A a B c C d D e E f m g S = f B = g

Split A Red-Black Tree b A a B c C d D e E S = f B = g

Split A Red-Black Tree b A a B c C d D e E S = f B = g S = join(e, E, S )

Split A Red-Black Tree b A a B c C d D S = f B = g S = join(e, E, S ) B = join( B, D, d)

Split A Red-Black Tree b A a B c C S = f B = g S = join(e, E, S ) B = join( B, D, d) S = join(c, C, S )

Split A Red-Black Tree b A a B S = f B = g S = join(e, E, S ) B = join( B, D, d) S = join(c, C, S ) B = join( B, B, b)

Split A Red-Black Tree A a S = f B = g S = join(e, E, S ) B = join( B, D, d) S = join(c, C, S ) B = join( B, B, b) S = join(a, A, S )

Complexity Of Split O(log n) See text.