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. The cost of these operations may become O(n) for a skewed Binary tree. If we make sure that height of the tree remains O(Logn) after every insertion and deletion, then we can guarantee an upper bound of O(Logn) for all these operations. The height of a Red-Black tree is always O(Logn) where n is the number of nodes in the tree.
Comparison with AVL Tree
The AVL trees are more balanced compared to Red-Black Trees, but they may cause more rotations during insertion and deletion. So if your application involves many frequent insertions and deletions, then Red Black trees should be preferred. And if the insertions and deletions are less frequent and search is a more frequent operation, then AVL tree should be preferred over Red-Black Tree.
How does a Red-Black Tree ensure balance?
A simple example to understand balancing is, a chain of 3 nodes is not possible in the red-black tree. We can try any combination of colours and see all of them violate Red-Black tree property.
A chain of 3 nodes is nodes is not possible in Red-Black Trees. Following are NOT Red-Black Trees 30 30 30 / \ / \ / \ 20 NIL 20 NIL 20 NIL / \ / \ / \ 10 NIL 10 NIL 10 NIL Violates Violates Violates Property 4. Property 4 Property 3 Following are different possible Red-Black Trees with above 3 keys 20 20 / \ / \ 10 30 10 30 / \ / \ / \ / \ NIL NIL NIL NIL NIL NIL NIL NIL
From the above examples, we get some idea how Red-Black trees ensure balance. Following is an important fact about balancing in Red-Black Trees.
Black Height of a Red-Black Tree :
Black height is number of black nodes on a path from a node to a leaf. Leaf nodes are also counted black nodes. From above properties 3 and 4, we can derive, a node of height h has black-height >= h/2.
Every Red Black Tree with n nodes has height <= 2Log2(n+1)
This can be proved using following facts:
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 atleast 7). This expression can also be written as k <= 2Log2(n+1)
2) From property 4 of Red-Black trees and above claim, we can say in a Red-Black Tree with n nodes, there is a root to leaf path with at-most Log2(n+1) black nodes.
3) From property 3 of Red-Black trees, we can claim that the number black nodes in a Red-Black tree is at least ⌊ n/2 ⌋ where n is the total number of nodes.
From above 2 points, we can conclude the fact that Red Black Tree with n nodes has height <= 2Log2(n+1)
In this post, we introduced Red-Black trees and discussed how balance is ensured. The hard part is to maintain balance when keys are added and removed. We will soon be discussing insertion and deletion operations in coming posts on the Red-Black tree.
1) Is it possible to have all black nodes in a Red-Black tree?
2) Draw a Red-Black Tree that is not an AVL tree structure wise?
Introduction to Algorithms 3rd Edition by Clifford Stein, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest
Video Lecture on Red-Black Tree by Tim Roughgarden
MIT Video Lecture on Red-Black Tree
MIT Lecture Notes on Red Black Tree
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
- Red-Black Tree | Set 2 (Insert)
- AVL Tree | Set 1 (Insertion)
- Red-Black Tree | Set 3 (Delete)
- B-Tree | Set 1 (Introduction)
- AVL Tree | Set 2 (Deletion)
- C++ Program to implement Symbol Table
- Iterative Preorder Traversal of an N-ary Tree
- Flatten a binary tree into linked list | Set-2
- Iterative Segment Tree (Range Maximum Query with Node Update)
- Print the DFS traversal step-wise (Backtracking also)