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Red-Black Tree
SWE2016-44
Example of a Red-Black Tree
2
Red-Black Tree
Red-Black Tree is a self-balancing Binary Search Tree (BST) where every node follows following rules:
3
1) Every node has a color either red or black
Red-Black Tree
Red-Black Tree is a self-balancing Binary Search Tree (BST) where every node follows following rules:
4
1) Every node has a color either red or black
2) Root of tree is always black
Red-Black Tree
Red-Black Tree is a self-balancing Binary Search Tree (BST) where every node follows following rules:
5
1) Every node has a color either red or black
2) Root of tree is always black
3) There are no two adjacent red nodes (A red node cannot have a red parent or red child)
Red-Black Tree
Red-Black Tree is a self-balancing Binary Search Tree (BST) where every node follows following rules:
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1) Every node has a color either red or black
2) Root of tree is always black
3) There are no two adjacent red nodes (A red node cannot have a red parent or red child)
4) Every path from root to a NULL node has same number of black nodes
Why Red-Black Trees?
Most of the BST operations (e.g., search, max, min, insert, delete.. etc) take O(h) time where h is the height of the BST.
→ O(n) for a skewed Binary tree.
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Why Red-Black Trees?
Most of the BST operations (e.g., search, max, min, insert, delete.. etc) take O(h) time where h is the height of the BST.
→ O(n) for a skewed Binary tree.
→ Since a Red-Black Tree ensures almost balanced, the height of the tree remains O(Log n) after every insertion and deletion.
→ Red-Black Tree is not always a balanced tree.
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How does a Red-Black Tree ensure balance?
We can try any combination of colors and see all of them violate Red-Black tree property.
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How does a Red-Black Tree ensure balance?
We can try any combination of colors and see all of them violate Red-Black tree property.
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Black Height of a Red-Black Tree
Black height is number of black nodes on a path from root to a leaf. Leaf nodes are also counted black nodes.
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Black Height of a Red-Black Tree
Black height is number of black nodes on a path from root to a leaf. Leaf nodes are also counted black nodes.
From properties 3 (no two adjacent red nodes) and 4 (same number of black nodes),
Black-height >= h/2.
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Height of a Red-Black Tree
Every Red Black Tree with n nodes has Height <= 2Log2(n+1).
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Height of a Red-Black Tree
Every Red Black Tree with n nodes has Height <= 2Log2(n+1).
Proof)
1) For a general Binary Tree, let k be the minimum number of nodes on all root to NULL paths, then n >= 2k – 1 (Ex. If k is 3, then n is at least 7). That is, k <= Log2 (n+1).
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Height of a Red-Black Tree
Every Red Black Tree with n nodes has Height <= 2Log2(n+1).
Proof)
1) For a general Binary Tree, let k be the minimum number of nodes on all root to NULL paths, then n >= 2k – 1 (Ex. If k is 3, then n is at least 7). That is, k <= Log2 (n+1).
2) From 1) and property 4, there is a root to leaf path with at most Log2 (n+1) black nodes: h' <= Log2 (n+1)
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Height of a Red-Black Tree
Every Red Black Tree with n nodes has Height <= 2Log2(n+1).
Proof)
1) For a general Binary Tree, let k be the minimum number of nodes on all root to NULL paths, then n >= 2k – 1 (Ex. If k is 3, then n is at least 7). That is, k <= Log2 (n+1).
2) From 1) and property 4, there is a root to leaf path with at most Log2 (n+1) black nodes: h' <= Log2 (n+1)
3) Black-height is at least h/2: h' >= h/2
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Height of a Red-Black Tree
Every Red Black Tree with n nodes has Height <= 2Log2(n+1).
Proof)
1) For a general Binary Tree, let k be the minimum number of nodes on all root to NULL paths, then n >= 2k – 1 (Ex. If k is 3, then n is at least 7). That is, k <= Log2 (n+1).
2) From 1) and property 4, there is a root to leaf path with at most Log2 (n+1) black nodes: h' <= Log2 (n+1)
3) Black-height is at least h/2: h' >= h/2
4) From 2) and 3), h <= 2Log2 (n+1)17
Insertion
The goal of the insert operation is to insert key K into tree T, maintaining T’s red-black tree properties
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Insertion
If T is a non-empty tree, then we do the following:
1. Use the BST insert algorithm to add K to the tree
2. Color the node containing K red
3. Restore red-black tree properties (if necessary)
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Insertion
To restore the violated property, we use:
1. Recoloring
2. Rotation (Left, Right, Double)
We try recoloring first, if recoloring doesn’t work, then we go for rotation.
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Insertion
1) Perform standard BST insertion and make the color of newly inserted nodes as RED.
2) If x is root, change color of x as BLACK (Black height of complete tree increases by 1).
3) Do following if color of x’s parent is not BLACK and x is not root.
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Insertion
a. If x’s uncle is RED (Grand parent must have been black from property 4)I. Change color of parent and uncle as BLACK.
II. color of grand parent as RED.
III. Change x = x’s grandparent, repeat steps 2 and 3 for new x.
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Insertion
b. If x’s uncle is BLACK, then there can be four configurations for x, x’s parent (p) and x’s grandparent (g)I. Left Left Case (p is left child of g and x is left child of p)
II. Left Right Case (p is left child of g and x is right child of p)
III. Right Right Case (Mirror of case i)
IV. Right Left Case (Mirror of case ii)
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Insertion
I. Left Left Case (p is left child of g and x is left child of p)
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rotateRight(root, g)
swap(p→color, g→color)
Insertion
II. Left Right Case (p is left child of g and x is right child of p)
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rotateLeft(root, p)
rotateRight(root, g)
swap(x→color, g→color)
Insertion
III. Right Right Case (Mirror of case i)
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rotateLeft(root, g)
swap(p→color, g→color)
Insertion
IV. Right Left Case (Mirror of case ii)
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rotateRight(root, p)
rotateLeft(root, g)
swap(x→color, g→color)
Insertion Analysis
• Go up the tree performing Case 3-a), which only recolors nodes.
• If Case 3-b) occurs, perform 1 or 2 rotations, and terminate.
→ Running time: O(log n) with O(1) rotations.
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Example of Insertion
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11
1
14
2
7
15
Example of Insertion
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11
1
14
2
7
15
Insert 1111
Example of Insertion
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11
1
14
2
7
15
Insert 1111
Example of Insertion
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11
1
14
2
7
15
Insert 1
1
11
Example of Insertion
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11
1
14
2
7
15
Insert 14
1
11
14
Example of Insertion
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11
1
14
2
7
15
Insert 2
1
11
14
2
Example of Insertion
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11
1
14
2
7
15
Insert 2
1
11
14
2
If X’s uncle is RED and X’s parent is not BLACK,
change color of parent and uncle as BLACK
Example of Insertion
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11
1
14
2
7
15
Insert 2
1
11
14
2
Example of Insertion
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11
1
14
2
7
15
Insert 2
1
11
14
2
Color of Grand parent as RED
Example of Insertion
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11
1
14
2
7
15
Insert 2
1
11
14
2
Example of Insertion
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11
1
14
2
7
15
Insert 2
1
11
14
2
As 11 is a root node, change its color to black
Example of Insertion
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11
1
14
2
7
15
Insert 2
1
11
14
2
Example of Insertion
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11
1
14
2
7
15
Insert 7
1
11
14
2
7
Example of Insertion
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11
1
14
2
7
15
Insert 7
1
11
14
2
7
Left rotate(2) and recolor nodes
Example of Insertion
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11
1
14
2
7
15
Insert 7
2
11
14
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Example of Insertion
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11
1
14
2
7
15
Insert 7
2
11
14
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Example of Insertion
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11
1
14
2
7
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Insert 15
2
11
14
71 15
Exercise
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Insert 3, 1, 5, 7, 6, 8, 9, 10
Exercise
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Insert 3, 1, 5, 7, 6, 8, 9, 10
3
6
8
51 97
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Deletion
Like Insertion, recoloring and rotations are used to maintain the Red-Black properties.
In delete operation, we check color of sibling to decide the appropriate case.
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Deletion Algorithm
1. Perform standard BST delete- A node which is either leaf or has only one child is deleted
- For an internal node: Copy the successor → call delete for successor
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Deletion Algorithm
1. Perform standard BST delete- A node which is either leaf or has only one child is deleted
- For an internal node: Copy the successor → call delete for successor
2. Handle cases where a node is leaf or has one child
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Deletion Algorithm
1. Perform standard BST delete- A node which is either leaf or has only one child is deleted
- For an internal node: Copy the successor → call delete for successor
2. Handle cases where a node is leaf or has one child
3. Let v be the node to be deleted and u be the child that replaces v- Note that u is NULL when v is a leaf and color of NULL is considered
as Black
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Deletion Algorithm
a) If Either u or v is RED- Mark the replaced child as black
b) If Both u and v are BLACK- Color u as a double black
- If u is root, make it single black and return52
v
u
v
uor → u
v
u→ u
a) If Either u or v is RED
b) If Both u and v are BLACK
Deletion Algorithm
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30
20 40
10
30
10 40v
u
Delete 20
30
20 40
30
NULL 40vDelete 20
50NULL NULL 50u
u
u
Deletion Algorithm
b) If Both u and v are BLACK- Let sibling of node be s
- Case 1: If s is black and at least one of s’s children is RED
- Case 2: If s is black and its both children are BLACK
- Case 3: If s is RED
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Deletion Algorithm
b) If Both u and v are BLACK- Let sibling of node be s
- Case 1: If s is black and at least one of s’s children is RED
- Let the red child of s be r
I. Left Left Case (Mirror of III)
II. Left Right Case (Mirror of IV)
III. Right Right Case
IV. Right Left Case
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30
NULL 40
50
su
r
p
Deletion Algorithm
b) If Both u and v are BLACK- Let sibling of node be s
- Case 1: If s is black and at least one of s’s children is RED
III. Right Right Case (r → outer node)
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30
20 40v
5035u
Delete 20
Deletion Algorithm
b) If Both u and v are BLACK- Let sibling of node be s
- Case 1: If s is black and at least one of s’s children is RED
III. Right Right Case (r → outer node)
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30
20 40v
5035
30
40u
5035
NULL
u
Delete 20
s
r
p
Deletion Algorithm
b) If Both u and v are BLACK- Let sibling of node be s
- Case 1: If s is black and at least one of s’s children is RED
III. Right Right Case (r → outer node)
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30
20 40v
5035
30
40u
5035
NULL
u
Delete 20
s
r
p Rotate Left (p)
Color exchange
(p&r→s, s→p)
Deletion Algorithm
b) If Both u and v are BLACK- Let sibling of node be s
- Case 1: If s is black and at least one of s’s children is RED
III. Right Right Case (r → outer node)
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30
20 40v
5035
30
40u
5035
NULL
u
40
50
35
30
Delete 20
s
r
p Rotate Left (p)
Color exchange
(p&r→s, s→p)
Deletion Algorithm
b) If Both u and v are BLACK- Let sibling of node be s
- Case 1: If s is black and at least one of s’s children is RED
IV. Right Left Case
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30
20 40v
35u
Delete 20
Deletion Algorithm
b) If Both u and v are BLACK- Let sibling of node be s
- Case 1: If s is black and at least one of s’s children is RED
IV. Right Left Case
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30
20 40v
35
30
40u
35
NULL
u
Delete 20
s
r
p
Deletion Algorithm
b) If Both u and v are BLACK- Let sibling of node be s
- Case 1: If s is black and at least one of s’s children is RED
IV. Right Left Case
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30
20 40v
35
30
40u
35
NULL
u
Delete 20
s
r
p Rotate Right (s)
Color exchange (r, s)
Deletion Algorithm
b) If Both u and v are BLACK- Let sibling of node be s
- Case 1: If s is black and at least one of s’s children is RED
IV. Right Left Case
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30
20 40v
35
30
40u
35
NULL
u
Delete 20
s
r
p 30
35u
40
NULL r
s
pRotate Right (s)
Color exchange (r, s)
Deletion Algorithm
b) If Both u and v are BLACK- Let sibling of node be s
- Case 1: If s is black and at least one of s’s children is RED
IV. Right Left Case
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30
20 40v
35
30
40u
35
NULL
u
Delete 20
s
r
p 30
35u
40
NULL s
r
pRotate Right (s)
Color exchange (r, s)
Same as case III
Deletion Algorithm
b) If Both u and v are BLACK- Let sibling of node be s
- Case 1: If s is black and at least one of s’s children is RED
IV. Right Left Case
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30
20 40v
35
30
40u
35
NULL
u
Delete 20
s
r
p 30
35u
40
NULL s
r
pRotate Right (s)
Color exchange (r, s)
Same as case III
Rotate Left (p)
Color exchange
(p&r→s, s→p)
Deletion Algorithm
b) If Both u and v are BLACK- Let sibling of node be s
- Case 1: If s is black and at least one of s’s children is RED
IV. Right Left Case
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30
20 40v
35
30
40u
35
NULL
u
Delete 20
s
r
p 30
35u
40
NULL s
r
pRotate Right (s)
Color exchange (r, s) 35
30 40
Same as case III
Rotate Left (p)
Color exchange
(p&r→s, s→p)
Deletion Algorithm
b) If Both u and v are BLACK- Let sibling of node be s
- Case 2: If s is black and its both children are BLACK1) Recolor p to BLACK and s to RED
2) If parent is BLACK, recur for parent
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Deletion Algorithm
b) If Both u and v are BLACK- Let sibling of node be s
- Case 2: If s is black and its both children are BLACK1) Recolor p to BLACK and s to RED
2) If parent is BLACK, recur for parent
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30
20 40v
u
Delete 20
Deletion Algorithm
b) If Both u and v are BLACK- Let sibling of node be s
- Case 2: If s is black and its both children are BLACK1) Recolor p to BLACK and s to RED
2) If parent is BLACK, recur for parent
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30
20 40v
u
Delete 20
30
40u NULL s
p
Deletion Algorithm
b) If Both u and v are BLACK- Let sibling of node be s
- Case 2: If s is black and its both children are BLACK1) Recolor p to BLACK and s to RED
2) If parent is BLACK, recur for parent
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30
20 40v
u
Delete 20
30
40u NULL s
ps → RED
p → DB
30
40NULL
Deletion Algorithm
b) If Both u and v are BLACK- Let sibling of node be s
- Case 2: If s is black and its both children are BLACK1) Recolor p to BLACK and s to RED
2) If parent is BLACK, recur for parent
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30
20 40v
u
Delete 20
30
40u NULL s
ps → RED
p → BLACK
30
40NULL
Deletion Algorithm
b) If Both u and v are BLACK- Let sibling of node be s
- Case 3: If s is RED1) Rotate p, Recolor p to RED and s to BLACK
2) Case I or Case II
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Deletion Algorithm
b) If Both u and v are BLACK- Let sibling of node be s
- Case 3: If s is RED1) Rotate p, Recolor p to RED and s to BLACK
2) Case I or Case II
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Delete 20
30
20 40v
5035u
Deletion Algorithm
b) If Both u and v are BLACK- Let sibling of node be s
- Case 3: If s is RED1) Rotate p, Recolor p to RED and s to BLACK
2) Case I or Case II
74
Delete 20
30
20 40v
5035u
30
NULL 40u
5035
s
p
Deletion Algorithm
b) If Both u and v are BLACK- Let sibling of node be s
- Case 3: If s is RED1) Rotate p, Recolor p to RED and s to BLACK
2) Case I or Case II
75
Delete 20
Rotate
p → RED
s → BLACK
30
20 40v
5035u
30
NULL 40u
5035
s
p 40
50
35
30
NULL
Deletion Algorithm
b) If Both u and v are BLACK- Let sibling of node be s
- Case 3: If s is RED1) Rotate p, Recolor p to RED and s to BLACK
2) Case I or Case II
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Delete 20
Rotate
p → RED
s → BLACK
30
20 40v
5035u
30
NULL 40u
5035
s
p 40
50
35
30
NULLu
p
s
Deletion Algorithm
b) If Both u and v are BLACK- Let sibling of node be s
- Case 3: If s is RED1) Rotate p, Recolor p to RED and s to BLACK
2) Case I or Case II
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Delete 20
Rotate
p → RED
s → BLACK
30
20 40v
5035u
30
NULL 40u
5035
s
p 40
50
35
30
NULLu
p
s
Case 2
Deletion Algorithm
b) If Both u and v are BLACK- Let sibling of node be s
- Case 3: If s is RED1) Rotate p, Recolor p to RED and s to BLACK
2) Case I or Case II
78
Delete 20
Rotate
p → RED
s → BLACK
30
20 40v
5035u
30
NULL 40u
5035
s
p 40
50
35
30
NULLu
p
s
Case 2 40
50
35
30
NULLu
p
s
Deletion Analysis
• Case 2 is the only case in which more iterations occur.
→ u moves up 1 level. Hence, O(log n) iterations.
• Each of cases 1 and 3 has at most ≤ 3 rotations in all
→ Running time: O(log n)
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Example of Deletion (1)
80
Example of Deletion (2)
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Example of Deletion (3)
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Example of Deletion (4)
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Example of Deletion (5)
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Example of Deletion (6)
85
Example of Deletion (7)
86
Example of Deletion (8)
87
Example of Deletion (9)
88
Example of Deletion (10)
89
Reference
• Charles Leiserson and Piotr Indyk, “Introduction to Algorithms”, September 29, 2004
• https://www.geeksforgeeks.org
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