1. Balanced Trees Balanced trees have height O(lg n).
Height-balanced trees
At each node, height of left and right subtrees are “close”.
e.g. AVL trees, B-trees, red-black trees, splay trees
Weight-balanced trees
At each node, number of nodes in left and right subtrees
are “close”.
Search, Predecessor, Successor, Minimum, Maximum
O(lg n) time.
Insert and Delete may have to rebalance the tree.

2. Rebalancing Heuristic -- Rotation
Right-Rotate(T, y) y x y C C
Left-Rotate(T, x) x A A B B
Preserves the inorder key sequence.
Rotation
Takes O(1) time.
To be used by insertion and deletion on a red-black tree.

3. An Example of Rotation 7 6 2 3 4 x y 7 6 2 3 4 11 9 9 18 19 20 22 1217 14 19 20 22 12 17 14
Left-Rotate(x) 18 11 y x

4. Red-Black Trees A “balanced” binary search tree with height (lg n).
Basic dynamic-set operations:
Search, Predecessor, Successor, Minimum
Maximum, Insert, Delete
all take O(lg n) time.

5. Node of a Red-Black Tree
color
key
left
right
parent
NIL as pointers to external nodes (leaves) of the tree.
Key-bearing nodes as internal nodes of the tree.

6. Red-Black Properties 12
internal
node 8 14
NILl 4 9 15
NILl
NILl
NILl
NILl
NILl 5
external node (requiring
no extra storage)
NILl
NILl
1. Every node is either red or black.
No path is more than twice
as long as any other.
2. The root is black.
3. Every leaf (NIL) is black.
4. If a node is red, then both of its children are black.
5. Every simple path from the root to a descendant leaf
contains the same number of black nodes.

7. Sentinel (Save Storage)
parent 12 8 14 4 9 15 5
nil(T)

8. All Nodes Are Black A complete binary tree! 12 8 14 4 9 13 20
NILl
NILl
NILl
NILl
NILl
NILl
NILl
NILl
Red nodes may be seen as “fill-ins” to a complete binary search tree.

9. Black-Height
The black-height of a node x, denoted bh(x), is the number
of black nodes on any path from x (excluded) to a leaf.
bh = 3 17 2 1
bh = 2 14 21 10 23 19 16 7 20
NILl
NILl 12 15
NILl
NILl 3
NILl
NILl
NILl
NILl
NILl
NILl
NILl
A node at height h has black-height ≥ h/2.
NILl
NILl

10. #Internal Nodes vs Black Height
Claim
Subtree rooted at node x contains  2 – 1 internal nodes.
bh(x)
Proof By induction on the height h of x.
h = 0 xx
2 – 1 = 0 internal node.
h > 0 x x or
≥bh(x)–1
bh(x)–1
black height
height ≤ h – 1
≥ 2 – 1
bh(x)–1
≥ 2 – 1
#internal nodes
(by induction)
≥ 2 – 1
2 – – = 2 – 1
bh(x)
bh(x)–1
internal node total: ≥

11. #Internal Nodes vs Height
Lemma
A red-black tree with n internal nodes has height at
most 2lg (n+1).
Proof The root of a red-black tree of height h has black-height  h/2.
This is because on any path down from the root a red node is always
followed by a black node (but not necessarily vice versa).
h/2
By the claim there are at least – 1 internal nodes in the tree.
Therefore n  – 1 and h  2 lg(n+1).
Corollary
Search, Minimum, Maximum, Successor, and Predecessor
can be done in O(lg n) time.