Insertion in an AVL Tree
AVL tree is a self-balancing Binary Search Tree (BST) where the difference between heights of left and right subtrees cannot be more than one for all nodes.
Example of AVL Tree:
The above tree is AVL because the differences between the heights of left and right subtrees for every node are less than or equal to 1.
Example of a Tree that is NOT an AVL Tree:
The above tree is not AVL because the differences between the heights of the left and right subtrees for 8 and 12 are greater than 1.
Why AVL 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 the height of the tree remains O(log(n)) after every insertion and deletion, then we can guarantee an upper bound of O(log(n)) for all these operations. The height of an AVL tree is always O(log(n)) where n is the number of nodes in the tree.
Insertion in AVL Tree:
To make sure that the given tree remains AVL after every insertion, we must augment the standard BST insert operation to perform some re-balancing.
Following are two basic operations that can be performed to balance a BST without violating the BST property (keys(left) < key(root) < keys(right)).
- Left Rotation
- Right Rotation
T1, T2 and T3 are subtrees of the tree, rooted with y (on the left side) or x (on the right side) y x / \ Right Rotation / \ x T3 - - - - - - - > T1 y / \ < - - - - - - - / \ T1 T2 Left Rotation T2 T3 Keys in both of the above trees follow the following order keys(T1) < key(x) < keys(T2) < key(y) < keys(T3) So BST property is not violated anywhere.
Steps to follow for insertion:
Let the newly inserted node be w
- Perform standard BST insert for w.
- Starting from w, travel up and find the first unbalanced node. Let z be the first unbalanced node, y be the child of z that comes on the path from w to z and x be the grandchild of z that comes on the path from w to z.
- Re-balance the tree by performing appropriate rotations on the subtree rooted with z. There can be 4 possible cases that need to be handled as x, y and z can be arranged in 4 ways.
- Following are the possible 4 arrangements:
- y is the left child of z and x is the left child of y (Left Left Case)
- y is the left child of z and x is the right child of y (Left Right Case)
- y is the right child of z and x is the right child of y (Right Right Case)
- y is the right child of z and x is the left child of y (Right Left Case)
Following are the operations to be performed in above mentioned 4 cases. In all of the cases, we only need to re-balance the subtree rooted with z and the complete tree becomes balanced as the height of the subtree (After appropriate rotations) rooted with z becomes the same as it was before insertion.
1. Left Left Case
T1, T2, T3 and T4 are subtrees. z y / \ / \ y T4 Right Rotate (z) x z / \ - - - - - - - - -> / \ / \ x T3 T1 T2 T3 T4 / \ T1 T2
2. Left Right Case
z z x / \ / \ / \ y T4 Left Rotate (y) x T4 Right Rotate(z) y z / \ - - - - - - - - -> / \ - - - - - - - -> / \ / \ T1 x y T3 T1 T2 T3 T4 / \ / \ T2 T3 T1 T2
3. Right Right Case
z y / \ / \ T1 y Left Rotate(z) z x / \ - - - - - - - -> / \ / \ T2 x T1 T2 T3 T4 / \ T3 T4
4. Right Left Case
z z x / \ / \ / \ T1 y Right Rotate (y) T1 x Left Rotate(z) z y / \ - - - - - - - - -> / \ - - - - - - - -> / \ / \ x T4 T2 y T1 T2 T3 T4 / \ / \ T2 T3 T3 T4
Illustration of Insertion at AVL Tree
The idea is to use recursive BST insert, after insertion, we get pointers to all ancestors one by one in a bottom-up manner. So we don’t need a parent pointer to travel up. The recursive code itself travels up and visits all the ancestors of the newly inserted node.
Follow the steps mentioned below to implement the idea:
- Perform the normal BST insertion.
- The current node must be one of the ancestors of the newly inserted node. Update the height of the current node.
- Get the balance factor (left subtree height – right subtree height) of the current node.
- If the balance factor is greater than 1, then the current node is unbalanced and we are either in the Left Left case or left Right case. To check whether it is left left case or not, compare the newly inserted key with the key in the left subtree root.
- If the balance factor is less than -1, then the current node is unbalanced and we are either in the Right Right case or Right-Left case. To check whether it is the Right Right case or not, compare the newly inserted key with the key in the right subtree root.
Below is the implementation of the above approach:
Preorder traversal of the constructed AVL tree is 30 20 10 25 40 50
Time Complexity: O(n*log(n)), For Insertion
Auxiliary Space: O(1)
The rotation operations (left and right rotate) take constant time as only a few pointers are being changed there. Updating the height and getting the balance factor also takes constant time. So the time complexity of the AVL insert remains the same as the BST insert which is O(h) where h is the height of the tree. Since the AVL tree is balanced, the height is O(Logn). So time complexity of AVL insert is O(Logn).
Comparison with Red Black Tree:
The AVL tree and other self-balancing search trees like Red Black are useful to get all basic operations done in O(log n) time. 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 the more frequent operation, then the AVL tree should be preferred over Red Black Tree.
Following is the post for deletion in AVL Tree:
AVL Tree | Set 2 (Deletion)