CSE 2331

Red-Black Trees

6.1

CSE 2331

Red-Black Trees

Definition. A red-black tree is a binary search tree with the

following properties:

• Every node is either red or black;

• The root is black;

• Every leaf is NIL and is black;

• If a node is red, then both its children are black;

• For each node, all simple paths from the node to descendant

leaves contain the same number of black nodes.

(Note: Every node in a binary tree is either a leaf or has BOTH a

left AND right child.)

6.2

CSE 2331

Red-Black Tree: Example

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6.3

CSE 2331

Red-Black Tree: Example

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6.4

CSE 2331

NOT a Red-Black Tree

This tree is NOT a red-black tree. Why not?

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6.5

CSE 2331

Red-Black Tree Exercise

Color the following tree so that it is a red-black tree:

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6.6

CSE 2331

Red-Black Tree Height

Definition. A red-black tree satisfies the following properties:

• Every node is either red or black;

• The root is black;

• Every leaf is NIL and is black;

• If a node is red, then both its children are black;

• For each node, all simple paths from the node to descendant

leaves contain the same number of black nodes.

Theorem. A red-black tree with n internal nodes has height at

most 2 log2(n + 1).

6.7

CSE 2331

Complete Binary Tree Size

Theorem 1. A complete binary search tree of height h has

2h+1 − 1 nodes.

Proof. Level i has 2i nodes (i = 0, 1, . . . , h).

1 + 2 + 22 + 23 + . . . + 2h = 2h+1 − 1.

6.8

CSE 2331

Red-Black Tree Size

Theorem 2. A red-black tree of height h has at least 2⌈h/2⌉ − 1

internal nodes.

Proof. (By Dr. Y. Wang.)

Let T be a red-black tree of height h.

Remove the leaves of T forming a tree T ′ of height h− 1.

Let r be the root of T ′.

Since no child of a red node is red and r is black, the longest path

from r to a leaf has at least ⌈h/2⌉ black nodes.

Since every path from r to a leaf has the same number of black

nodes, every path from r to a leaf has at least ⌈h/2⌉ black nodes.

Thus, T ′ contains a complete binary tree of height at least ⌈h/2⌉−1.

(Note: height = Number of EDGES on longest path from root to

leaf.)

By Theorem 1, tree T ′ has at least 2⌈h/2⌉−1+1 − 1 nodes so tree T

has at least 2⌈h/2⌉ − 1 internal nodes.

6.9

CSE 2331

Red-Black Tree Height

Theorem. A red-black tree with n internal nodes has height at

most 2 log2(n + 1).

Proof. Let h be the height of a red-black tree with n nodes.

By Theorem 2, n ≥ 2⌈h/2⌉ − 1.

Thus, log2(n + 1) ≥ ⌈h/2⌉ ≥ h/2 so h ≤ 2 log2(n + 1).

6.10

CSE 2331

Red-Black Tree: Insert

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6.11

CSE 2331

Red-Black Tree Insert

function RBLocateParent(T , z)

/* Return future parent of z in tree */

1 y ← NIL;

2 x← T.root;

3 while (x is not a leaf) do

4 y ← x;

5 if (z.key < x.key) then x← x.left;

6 else x← x.right;

7 end

8 return (y);

6.12

CSE 2331

Red-Black Tree Insert

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17 25function RBTreeInsert(T , z)

1 y ← RBLocateParent(T, z);

2 z.parent← y;

3 if (y = NIL) then T.root← z; /* tree T was empty*/

4 else if (z.key < y.key) then y.left← z;

5 else y.right← z;

6 z.left← leaf;

7 z.right← leaf;

8 z.color← Red;

9 RBInsertFixup(T ,z);

6.13

CSE 2331

Insert Fixup: Case I

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If the parent and “uncle” of z are Red:

• Color the parent of z Black;

• Color the uncle of z Black;

• Color the grandparent of z Red;

• Repeat on the grandparent of z.

6.14

CSE 2331

Red-Black Tree Insert Fixup: Case I

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6.15

CSE 2331

Sibling

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function Sibling(x)

/* Return sibling of x */

1 if (x.parent = NIL) then error “Root has no siblings.”;

2 p← x.parent;

3 if (p.left = x) then return (p.right);

4 else return (p.left);

6.16

CSE 2331

Red-Black Tree Insert Fixup: Case I

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function RBInsertFixupA(T , alters z)

1 while (z 6= T.root) and (z.parent.color 6= Black) do

2 y ← Sibling(z.parent);

3 if (y.color = Black) then return;

4 z.parent.color← Black;

5 y.color← Black;

6 z ← z.parent.parent;

7 z.color← Red;

8 end

6.17

CSE 2331

Insert Fixup: Case III

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If the parent of z is red and its “uncle” is black:If z is a left child and its parent is a left child:

• Right Rotate on the grandparent of z;

• Color the parent of z Black;

• Color the sibling of z red.

6.18

CSE 2331

Red-Black Tree Insert Fixup: Case III

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6.19

CSE 2331

Red-Black Tree Insert Fixup: Case III

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function RBInsertFixupC(T , alters z)

1 if (z = T.root) or (z.parent.color = Black) then return;

2 x← z.parent;

3 w ← x.parent;

4 if (z = x.left) and (x = w.left) then

5 RightRotate(T ,w);

6 x.color← Black;

7 w.color← Red;

8 else if (z = x.right) and (x = w.right) then

9 Handle same as above with “right” and “left” exchanged

10 . . .

6.20

CSE 2331

Insert Fixup: Case II

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If the parent x of z is red and its “uncle” is black:If z is a right child and its parent x is a left child:

• z ← x

• Left Rotate on x;

• Apply algorithm for Case III.

6.21

CSE 2331

Red-Black Tree Insert Fixup: Case II

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6.22

CSE 2331

Red-Black Tree Insert Fixup: Case II

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function RBInsertFixupB(T , alters z)

1 if (z = T.root) or (z.parent.color = Black) then return;

2 x← z.parent;

3 w ← x.parent;

4 if (z = x.right) and (x = w.left) then

5 z ← x;

6 LeftRotate(T ,x);

7 else if (z = x.left) and (x = w.right) then

8 Handle same as above with “right” and “left” exchanged

9 . . .

6.23

CSE 2331

Red-Black Tree Insert Fixup

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function RBInsertFixup(T , z)

1 RBInsertFixupA (T ,z);

2 RBInsertFixupB (T ,z);

3 RBInsertFixupC (T ,z);

4 T.root.color← Black;

6.24

CSE 2331

Red-Black Tree Insert Fixup

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6.25

CSE 2331

Running Time Analysis: RBInsertFixup

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function RBInsertFixup(T , z)

1 RBInsertFixupA (T ,z);

2 RBInsertFixupB (T ,z);

3 RBInsertFixupC (T ,z);

4 T.root.color← Black;

6.26