Red black tree

Algorithms

Sandeep Kumar Poonia
Head Of Dept. CS/IT
B.E., M.Tech., UGC-NET
LM-IAENG, LM-IACSIT,LM-CSTA, LM-AIRCC, LM-SCIEI, AM-UACEE

Binary search trees
● Binary search trees are an important data

structure for dynamic sets.
● Accomplish many dynamic-set operations in
O(h) time, where h=height of tree.
● we represent a binary tree by a linked data
structure in which each node is an object.
● T:root points to the root of tree T .

Binary search trees
● Each node contains the attributes
■ key (and possibly other satellite data).
■ left: points to left child.
■ right: points to right child.

■ p: points to parent. T.root.p = NIL.

Binary search trees
● Stored keys must satisfy the binary-search-

tree property.
■ If y is in left subtree of x, then y.key <= x.key.
■ If y is in right subtree of x, then y.key >= x.key.

Binary search trees

(a) A binary search tree on 6 nodes with height 2.
(b) A less efficient binary search tree with height 4 that
contains the same keys.

Binary Search Trees

A binary search tree

Not a binary search tree

Binary search trees
● The binary-search-tree property allows us to

print out all the keys in a binary search tree in
sorted order by a simple recursive algorithm,
called an inorder tree walk.
It takes ‚
time to
walk an n-node
binary search tree

Binary search trees
● A: prints elements in sorted (increasing) order

● This is called an inorder tree walk
■ Preorder tree walk: print root, then left, then right
■ Postorder tree walk: print left, then right, then root

Tree traversal
● Used to print out the data in a tree in a certain

order
● Pre-order traversal
■ Print the data at the root
■ Recursively print out all data in the left subtree
■ Recursively print out all data in the right subtree

Preorder, Postorder and Inorder
● Preorder traversal
■ node, left, right
■ prefix expression
○ ++a*bc*+*defg

Preorder, Postorder and Inorder
● Postorder traversal
■ left, right, node
■ postfix expression
○ abc*+de*f+g*+

● Inorder traversal
■ left, node, right.
■ infix expression
○ a+b*c+d*e+f*g

Binary search trees
the running time of
TREE-SEARCH is
O(h), where h is
the height of the
tree.

insert
● Proceed down the tree as you would with a find
● If X is found, do nothing (or update something)
● Otherwise, insert X at the last spot on the path traversed

● Time complexity = O(height of the tree)

delete
● When we delete a node, we need to consider

how we take care of the children of the deleted
node.
■ This has to be done such that the property of the

search tree is maintained.

delete
Three cases:
(1) the node is a leaf
■ Delete it immediately

(2) the node has one child
■ Adjust a pointer from the parent to bypass that node

delete
(3) the node has 2 children
■ replace the key of that node with the minimum element at the right

subtree
■ delete the minimum element
○ Has either no child or only right child because if it has a left child, that

left child would be smaller and would have been chosen. So invoke
case 1 or 2.

●

Time complexity = O(height of the tree)

Balanced Binary Search Trees
A binary search tree can implement any of the basic dynamic-set
operations in O(h) time. These operations are O(lgn) if tree is
“balanced”.
There has been lots of interest in developing algorithms to keep binary
search trees balanced, including
1st type: insert nodes as is done in the BST insert, then rebalance tree
 Red-Black trees
 AVL trees
 Splay trees
2nd type: allow more than one key per node of the search tree:
 2-3 trees
 2-3-4 trees
 B-trees

Red-Black Trees (RBT)
Red-Black tree: BST in which each node is colored red or black.

Constraints on the coloring and connection of nodes ensure that
no root to leaf path is more than twice as long as any other, so
tree is approximately balanced.
Each RBT node contains fields left, right, parent, color, and key.

L
E
F
T

PARENT
KEY
COLOR

R
I
G
H
T

Red-Black Trees
● Red-black trees:
■ Binary search trees augmented with node color
■ Operations designed to guarantee that the height

h = O(lg n)
● First: describe the properties of red-black trees
● Then: prove that these guarantee h = O(lg n)
● Finally: describe operations on red-black trees

Red-Black Properties
● The red-black properties:

1. Every node is either red or black
2. Every leaf (NULL pointer) is black
○ Note: this means every “real” node has 2 children

3. If a node is red, both children are black
○ Note: can’t have 2 consecutive reds on a path

4. Every path from node to descendent leaf contains
the same number of black nodes
5. The root is always black

Red-Black Trees
● Put example on board and verify properties:

1.
2.
3.
4.

Every node is either red or black
Every leaf (NULL pointer) is black
If a node is red, both children are black
Every path from node to descendent leaf contains
the same number of black nodes
● black-height: # black nodes on path to leaf
■ Label example with h and bh values

Black Height bh(x)
Black-height of a node x: bh(x) is the number of black nodes (including
the NIL leaf) on the path from x to a leaf, not counting x itself.
2 20
2 18
1 17

0

0

1 22

1 19

0

Every node has a
black-height, bh(x).
For all NIL leaves,
bh(x) = 0.

1 21 1 25

0

0

0 0

0

For root x,
bh(x) = bh(T).

Height of a Red-black Tree
h=4
26 bh=2

● Example:
● Height of a node:
■ Number of edges in a
longest path to a leaf.
● Black-height of a node

17

h=1
bh=1

h=2 30
bh=1

bh(x) is the number of
black nodes on path
from x to leaf, not
counting x.
nil[T]

h=3
41 bh=2

h=1
bh=1
38

h=2
47 bh=1
h=1 50
bh=1

Height of Red-Black Trees
● What is the minimum black-height of a node
●
●
●
●

with height h?
A height-h node has black-height  h/2
Proof : By property 4,
h/2 nodes on the path from the node to a leaf
are red.
Hence
are black.

● The subtree rooted at any node x contains >= 2bh(x)_1

internal nodes.
● Proof :By induction height of x =0x is a leafbh(x)=0.
The subtree rooted at x has 0 internal nodes. 20 -1 = 0.
● Let the height of x be h and bh(x) = b.
● Any child of x has height h -1 and
● black-height either b (if the child is red) or
● b -1 (if the child is black)
● By induction each child has >= 2bh(x)-1-1 internal nodes
● Thus, the subtree rooted at x contains >= 2(2bh(x)-1-1)+1
●
= 2bh(x)-1(internal Nodes)

● Theorem: A red-black tree with n internal

nodes has height h  2 lg(n + 1)
● How do you suppose we’ll prove this?
● Proof: The subtree rooted at any node x contains
● >= 2bh(x)_1 internal nodes.
● A height-h node has black-height  h/2

n  2h/2 -1
n + 1 2h/2
lg(n+1)  h/2  h  2lg(n+1)

● Thus
●
●

RB Trees: Worst-Case Time
● So we’ve proved that a red-black tree has

O(lg n) height
● Corollary: These operations take O(lg n) time:
■ Minimum(), Maximum()
■ Successor(), Predecessor()
■ Search()

● Insert() and Delete():
■ Will also take O(lg n) time
■ But will need special care since they modify tree

Red-Black Trees: An Example
● Color this tree:

7
5

9
12

Red-black properties:
1.
2.
3.
4.
Every path from node to descendent leaf
contains the same number of black nodes
5.
The root is always black

Red-Black Trees:
The Problem With Insertion
● Insert 8
■ Where does it go?

7
5

9
12

1.
2.
3.
4.
5.


Red-Black Trees:
● Insert 8
■ What color

7
5

should it be?

1.
2.
3.
4.
5.


9
8

12

Red-Black Trees:
● Insert 11

7
5

9
8

1.
2.
3.
4.
5.


12

Red-Black Trees:
● Insert 11
■ What color?

7
5

9
8

12
11

1.
2.
3.
4.
5.


Red-Black Trees:
● Insert 11
■ What color?
○ Can’t be red! (#3)

7
5

9
8

12
11

1.
2.
3.
4.
5.


Red-Black Trees:
● Insert 11
■ What color?
○ Can’t be red! (#3)
○ Can’t be black! (#4)

1.
2.
3.
4.
5.

7
5


9
8

12
11

Red-Black Trees:
● Insert 11
■ What color?
○ Solution:
recolor the tree

1.
2.
3.
4.
5.

7
5


9
8

12
11

Red-Black Trees:
● Insert 10

7
5

9
8

12
11

1.
2.
3.
4.
5.


Red-Black Trees:
● Insert 10
■ What color?

7
5

9
8

12
11

1.
2.
3.
4.
5.


10

Red-Black Trees:
● Insert 10

7
5

■ What color?
○ A: no color! Tree
is too imbalanced
○ Must change tree structure
to allow recoloring
■ Goal: restructure tree in

O(lg n) time

9
8

12
11
10

RB Trees: Rotation
● Our basic operation for changing tree structure

is called rotation:
y
x

x

rightRotate(y)

C

A

y

leftRotate(x)

A

B

B

C

● Does rotation preserve inorder key ordering?
● What would the code for rightRotate()

actually do?

RB Trees: Rotation
y
x
A

x

rightRotate(y)

C

A

B

y
B

● A lot of pointer manipulation
■ x keeps its left child
■ y keeps its right child

■ x’s right child becomes y’s left child
■ x’s and y’s parents change

C

RB Trees: Rotation
y
x
A

x

leftRotate(x)
rightRotate(y)

C

A

B

y
B

● A lot of pointer manipulation
■ x keeps its left child
■ y keeps its right child

■ y’s right child becomes x’s left child
■ x’s and y’s parents change

C

RB Trees: Rotation
Left-Rotate (T, x)
1. y  right[x] // Set y.
2. right[x]  left[y] //Turn y’s left subtree into x’s right subtree.
3. if left[y]  nil[T ]
4.
then p[left[y]]  x
5. p[y]  p[x]
// Link x’s parent to y.
6. if p[x] == nil[T ]
7.
then root[T ]  y
8.
else if x == left[p[x]]
9.
then left[p[x]]  y
10.
else right[p[x]]  y
11. left[y]  x
// Put x on y’s left.
12. p[x]  y

RB Trees: Rotation
right-Rotate (T, y)
1. x  right[y]
2. left[y]  right[x]
3. If right[x]  nil[T ]
4.
then p[right[x]]  y
5. p[x]  p[y]
6. if p[y] == nil[T ]
7.
then root[T ]  x
8.
else if y == left[p[y]]
9.
then left[p[y]]  x
10.
else right[p[y]]  x
11. right[x]  y
12. p[y]  x

Rotation Example
● Rotate left about 9:
7

5

9
8

12
11

Rotation Example
● Rotate left about 9:
7

5

12
9
8

11

Red-Black Trees: Insertion
● Remember:

1. Insert nodes one at a time, and after every
Insertion balance the tree.
2. Every node inserted starts as a Red node.
3. Consult the cases, Every time two Red nodes
touch must rebalance at that point.
4. The root will always be Black.

Red-Black Trees: Insertion
● If we insert, what color to make the new node?

■ Red? Might violate property 3(If a node is red, both
children are black).

■ Black? Might violate property 4(Every path from node
to descendent leaf contains the same number of black nodes).

Insertion: the basic idea
■ Insert x into tree, color x red

■ Only r-b property 3 might be violated (if p[x] red)
○ If so, move violation up tree until a place is found where it
can be fixed
■ Total time will be O(lg n)

Reminder: Red-black Properties
1. Every node is either red or black

2. Every leaf (NULL pointer) is black
3. If a node is red, both children are black
4. Every path from node to descendent leaf


Insertion in RB Trees
● Insertion must preserve all red-black properties.

● Should an inserted node be colored Red? Black?
● Basic steps:
■ Use Tree-Insert from BST (slightly modified) to

insert a node x into T.
○ Procedure RB-Insert(x).

■ Color the node x red.

■ Fix the modified tree by re-coloring nodes and

performing rotation to preserve RB tree property.
○ Procedure RB-Insert-Fixup.

Algorithm: Insertion
We have detected a need for balance when z is red and his parent too.

• If z has a red uncle: colour the parent and uncle black, and
grandparent red.

z

Balanced trees: Red-black trees


grandparent red.
• If z is a left child and has a black uncle: colour the parent black and
the grandparent red, then rotateRight(z.parent.parent)



grandparent red.
right child and has black uncle: colour the parent black and
• If z is a left child and has a a black uncle, then rotateLeft(z.parent)
andgrandparent red, then rotateRight(z.parent.parent)
the


Let’s insert 4, 7, 12, 15, 3 and 5.

4

7
1
2

Double red violation!

It also shows it’s
unbalanced…


Let’s insert 4, 7, 12, 15, 3 and 5.

7
4

What should we do?

Nothing, no
double red.

3

12
5

15

Double red violation.
We can’t have a better
balance, and there is a
red uncle…


Insertion
RB-Insert(T, z)
1.
y  nil[T]
2.
x  root[T]
3.
while x  nil[T]
4.
do y  x
5.
if key[z] < key[x]
6.
then x  left[x]
7.
else x  right[x]
8.
p[z]  y
9.
if y = nil[T]
10.
then root[T]  z
11.
else if key[z] < key[y]
12.
then left[y]  z
13.
else right[y]  z

RB-Insert(T, z) Contd.
14. left[z]  nil[T]
15. right[z]  nil[T]
16. color[z]  RED
17. RB-Insert-Fixup (T, z)

How does it differ from the TreeInsert procedure of BSTs?
Which of the RB properties might
be violated?
Fix the violations by calling RBInsert-Fixup.

Insertion – Fixup
● Which property might be violated?
1. OK(Every node is either red or black)
2. If z is the root, then there’s a violation. Otherwise,

OK(The root is black)
3. OK(Every leaf (NIL) is black)
4. If z.p is red, there’s a violation: both z and z.p are
red(If a node is red, then both its children are black)
● OK(For each node, all simple paths from the node to descendant
leaves contain the same number of black nodes)
● Remove the violation by calling RB-INSERT-FIXUP:

Insertion – Fixup
● Problem: we may have one pair of consecutive

reds where we did the insertion.
● Solution: rotate it up the tree and away…

Three cases have to be handled…

Insertion – Fixup
RB-Insert-Fixup (T, z)
1.
while color[p[z]] == RED
2.
do if p[z] == left[p[p[z]]]
3.
then y  right[p[p[z]]]
4.
if color[y] == RED
5.
then color[p[z]]  BLACK // Case 1
6.
color[y]  BLACK
// Case 1
7.
color[p[p[z]]]  RED // Case 1
8.
z  p[p[z]]
// Case 1

Insertion – Fixup
RB-Insert-Fixup(T, z) (Contd.)
9.
else if z == right[p[z]] // color[y]  RED
10.
then z  p[z]
// Case 2
11.
LEFT-ROTATE(T, z)
// Case 2
12.
color[p[z]]  BLACK
// Case 3
13.
color[p[p[z]]]  RED
// Case 3
14.
RIGHT-ROTATE(T, p[p[z]]) // Case 3
15.
else (if p[z] = right[p[p[z]]])(same as 10-14
16.
with “right” and “left” exchanged)
17. color[root[T ]]  BLACK

Insertion – Fixup

A node z ’after insertion. Because both z and its parent z.p are red, a
violation of property 4 occurs. Since z’s uncle y is red, case 1 in the
code applies. We recolor nodes and move the pointer z up the tree

Insertion – Fixup

Once again, z and its parent are both red, but z’s uncle y
is black. Since z is the right child of z.p, case 2 applies.
We perform a left rotation,

Insertion – Fixup

Now, z is the left child of its parent, and case 3 applies.
Recoloring and right rotation yield the tree

Correctness
Loop invariant:
● At the start of each iteration of the while loop,
■ z is red.
■ If p[z] is the root, then p[z] is black.
■ There is at most one red-black violation:
○ Property 2: z is a red root, or
○ Property 4: z and p[z] are both red.

Correctness – Contd.
● Initialization: 

● Termination: The loop terminates only if p[z] is

black. Hence, property 4 is OK.
The last line ensures property 2 always holds.
● Maintenance: We drop out when z is the root (since
then p[z] is sentinel nil[T ], which is black). When we
start the loop body, the only violation is of property 4.
■ There are 6 cases, 3 of which are symmetric to the other 3.

We consider cases in which p[z] is a left child.
■ Let y be z’s uncle (p[z]’s sibling).

Case 1 – uncle y is red

p[p[z]]

new z
C

C

p[z]

y

A

D
z



A





B





D







B

z is a right child here.
Similar steps if z is a left child.





● p[p[z]] (z’s grandparent) must be black, since z and p[z] are both red and there

are no other violations of property 4.
● Make p[z] and y black

 now z and p[z] are not both red. But property 5

might now be violated.
● Make p[p[z]] red

 restores property 5.

● The next iteration has p[p[z]] as the new z (i.e., z moves up 2 levels).

Case 1 – uncle y is red

We take the same action whether z is a right child or z is a left child.
Each subtree has a black root, and each has the same black-height.
The code for case 1 changes the colors of some nodes, preserving property
5: all downward simple paths from a node to a leaf have the same number
of blacks. The while loop continues with node z’s grandparent z.p.p as
the new z. Any violation of property 4 can now occur only between the new
z, which is red, and its parent, if it is red as well.

Case 2 – y is black, z is a right child
C

C

p[z]

p[z]

 y

A

 y

B

z



z

B







A





● Left rotate around p[z], p[z] and z switch roles  now z is a left

child, and both z and p[z] are red.
● Takes us immediately to case 3.

Case 3 – y is black, z is a left child
B

C
p[z]

 y

B
z



A



A



C







● Make p[z] black and p[p[z]] red.
● Then right rotate on p[p[z]]. Ensures property 4 is maintained.
● No longer have 2 reds in a row.
● p[z] is now black  no more iterations.



Cases 2 and 3 of the procedure RB-INSERT-FIXUP

As in case 1, property 4 is violated in either case 2 or case 3 because z
and its parent z.p are both red. Each subtree has a black root , and each
has the same black-height. We transform case 2 into case 3 by a left
rotation, which preserves property 5: all downward simple paths from a
node to a leaf have the same number of blacks. Case 3 causes some
color changes and a right rotation, which also preserve property 5. The
while loop then terminates, because property 4 is satisfied: there are no
longer two red nodes in a row.

Algorithm Analysis
● O(lg n) time to get through RB-Insert up to the

call of RB-Insert-Fixup.
● Within RB-Insert-Fixup:
■ Each iteration takes O(1) time.

■ Each iteration but the last moves z up 2 levels.
■ O(lg n) levels  O(lg n) time.
■ Thus, insertion in a red-black tree takes O(lg n) time.
■ Note: there are at most 2 rotations overall.

Deletion
● Deletion, like insertion, should preserve all the

RB properties.
● The properties that may be violated depends on
the color of the deleted node.
■ Red – OK. Why?

■ Black?

● Steps:
■ Do regular BST deletion.
■ Fix any violations of RB properties that may result.

Deletion
RB-Delete(T, z)
1.
if left[z] == nil[T] or right[z] == nil[T]
2.
then y  z
3.
else y  TREE-SUCCESSOR(z)
4.
if left[y] == nil[T ]
5.
then x  left[y]
6.
else x  right[y]
7.
p[x]  p[y] // Do this, even if x is nil[T]

Deletion
RB-Delete (T, z) (Contd.)
8. if p[y] == nil[T ]
9.
then root[T ]  x
10. else if y == left[p[y]]
11.
then left[p[y]]  x
12.
else right[p[y]]  x
13. if y == z
14. then key[z]  key[y]
15. copy y’s satellite data into z
16. if color[y] == BLACK
17. then RB-Delete-Fixup(T, x)
18. return y

The node passed to
the fixup routine is the
lone child of the
spliced up node, or
the sentinel.

RB Properties Violation
● If the delete node is red?

Not a problem – no RB properties violated
● If y is black, we could have violations of redblack properties:
■ Prop. 1. OK.
■ Prop. 2. If y is the root and x is red, then the root has

become red.
■ Prop. 3. OK.
■ Prop. 4. Violation if p[y] and x are both red.
■ Prop. 5. Any path containing y now has 1 fewer black

RB Properties Violation
● Prop. 5. Any path containing y now has 1 fewer black

node.
■ Correct by giving x an “extra black.”
■ Add 1 to count of black nodes on paths containing x.
■ Now property 5 is OK, but property 1 is not.

■ x is either doubly black (if color[x] == BLACK) or red &

black (if color[x] == RED).
■ The attribute color[x] is still either RED or BLACK. No
new values for color attribute.
■ In other words, the extra blackness on a node is by virtue of
x pointing to the node.
● Remove the violations by calling RB-Delete-Fixup.

Deletion – Fixup
RB-Delete-Fixup(T, x)
1.
while x  root[T ] and color[x] == BLACK
2.
do if x == left[p[x]]
3.
then w  right[p[x]]
4.
if color[w] == RED
5.
then color[w]  BLACK
// Case 1
6.
color[p[x]]  RED
// Case 1
7.
LEFT-ROTATE(T, p[x])
// Case 1
8.
w  right[p[x]]
// Case 1

RB-Delete-Fixup(T, x) (Contd.)
/* x is still left[p[x]] */
9.
if color[left[w]] == BLACK and color[right[w]] == BLACK
10.
then color[w]  RED
// Case 2
11.
x  p[x]
// Case 2
12.
else if color[right[w]] == BLACK
13.
then color[left[w]]  BLACK
// Case 3
14.
color[w]  RED
// Case 3
15.
RIGHT-ROTATE(T,w)
// Case 3
16.
w  right[p[x]]
// Case 3
17.
color[w]  color[p[x]]
// Case 4
18.
color[p[x]]  BLACK
// Case 4
19.
color[right[w]]  BLACK
// Case 4
20.
LEFT-ROTATE(T, p[x])
// Case 4
21.
x  root[T ]
// Case 4
22.
else (same as then clause with “right” and “left” exchanged)
23. color[x]  BLACK

Deletion – Fixup
● Idea: Move the extra black up the tree until x points
to a red & black node  turn it into a black node,
● x points to the root  just remove the extra black, or
● We can do certain rotations and recolorings and

finish.
● Within the while loop:
■ x always points to a nonroot doubly black node.
■ w is x’s sibling.
■ w cannot be nil[T ], since that would violate property 5 at

p[x].
● 8 cases in all, 4 of which are symmetric to the other.

Case 1 – w is red
p[x]
B

w

x
A



D
B

D



x
C



E

 





A



E

new
w





C





● w must have black children.
● Make w black and p[x] red (because w is red p[x] couldn’t have

been red).
● Then left rotate on p[x].
● New sibling of x was a child of w before rotation  must be
black.
● Go immediately to case 2, 3, or 4.

Case 2 – w is black, both w’s children
p[x]
are black new x
c
B

w

x
A



B

D

A




C



E

 

c

D


C





E

 

● Take 1 black off x ( singly black) and off w ( red).
● Move that black to p[x].
● Do the next iteration with p[x] as the new x.
● If entered this case from case 1, then p[x] was red  new x is
red & black  color attribute of new x is RED  loop

terminates. Then new x is made black in the last line.



Case 3 – w is black, w’s left child is
c
red, w’s right child is black
c
B
w

x
A



B

x

D

A




C



E

 

new w



C



D




E


● Make w red and w’s left child black.
● Then right rotate on w.

● New sibling w of x is black with a red right child  case 4.



Case 4 – w is black, w’s right child is
red
c
B

w

x
A



D
B

D


C



c’

 

x
E





E

A



C









● Make w be p[x]’s color (c).
● Make p[x] black and w’s right child black.
● Then left rotate on p[x].

● Remove the extra black on x ( x is now singly black) without

violating any red-black properties.
● All done. Setting x to root causes the loop to terminate.

Red black tree

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