AVL TREE

       In computer science, an AVL tree is a self-balancing binary search tree, and it was the first such data structure to be invented. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. Lookup, insertion, and deletion all take O(log n) time in both the average and worst cases, where n is the number of nodes in the tree prior to the operation. Insertions and deletions may require the tree to be rebalanced by one or more tree rotations. 

Operations

Tree rotations

      Basic operations of an AVL tree involve carrying out the same actions as would be carried out on an unbalanced binary search tree, but modifications are followed by zero or more operations called tree rotations, which help to restore the height balance of the subtrees.

Searching

     Once a node has been found in a balanced tree, the next or previous nodes can be explored in amortized constant time. Some instances of exploring these "nearby" nodes require traversing up to 2×log(n) links (particularly when moving from the rightmost leaf of the root's left subtree to the leftmost leaf of the root's right subtree). However, exploring all n nodes of the tree in this manner would use each link exactly twice: one traversal to enter the subtree rooted at that node, another to leave that node's subtree after having explored it. And since there are n−1 links in any tree, the amortized cost is found to be 2×(n−1)/n, or approximately 2.

Insertion

Pictorial description of how rotations cause rebalancing tree, and then retracing one's steps toward the root updating the balance factor of the nodes. The numbered circles represent the nodes being balanced. The lettered triangles represent subtrees which are themselves balanced BST.

After inserting a node, it is necessary to check each of the node's ancestors for consistency with the rules of AVL. The balance factor is calculated as follows: balanceFactor = height(left-subtree) - height(right-subtree). For each node checked, if the balance factor remains −1, 0, or +1 then no rotations are necessary. However, if balance factor becomes less than -1 or greater than +1, the subtree rooted at this node is unbalanced. If insertions are performed serially, after each insertion, at most one of the following cases needs to be resolved to restore the entire tree to the rules of AVL.

There are four cases which need to be considered, of which two are symmetric to the other two. Let P be the root of the unbalanced subtree, with R and L denoting the right and left children of P respectively.

Right-Right case and Right-Left case:

·         If the balance factor of P is -2 then the right subtree outweighs the left subtree of the given node, and the balance factor of the right child (R) must be checked. The left rotation with P as the root is necessary.

o   If the balance factor of R is -1 (or in case of deletion also 0), a single left rotation(with P as the root) is needed (Right-Right case).

o   If the balance factor of R is +1, two different rotations are needed. The first rotation is a right rotation with R as the root. The second is a left rotation with P as the root (Right-Left case).

Left-Left case and Left-Right case:

·         If the balance factor of P is 2, then the left subtree outweighs the right subtree of the given node, and the balance factor of the left child (L) must be checked. The right rotation with P as the root is necessary.

o   If the balance factor of L is +1 (or in case of deletion also 0), a single right rotation(with P as the root) is needed (Left-Left case).

o   If the balance factor of L is -1, two different rotations are needed. The first rotation is a left rotation with L as the root. The second is a right rotation with P as the root (Left-Right case).

Deletion

          If the node is a leaf or has only one child, remove it. Otherwise, replace it with either the largest in its left sub tree (in order predecessor) or the smallest in its right sub tree (in order successor), and remove that node. The node that was found as a replacement has at most one sub tree. After deletion, retrace the path back up the tree (parent of the replacement) to the root, adjusting the balance factors as needed.
As with all binary trees, a node's in-order successor is the left-most child of its right subtree, and a node's in-order predecessor is the right-most child of its left subtree. In either case, this node will have zero or one children. Delete it according to one of the two simpler cases above.

In addition to the balancing described above for insertions, if the balance factor for the tree is 2 and that of the left subtree is 0, a right rotation must be performed on P. The mirror of this case is also necessary.
The retracing can stop if the balance factor becomes −1 or +1 indicating that the height of that subtree has remained unchanged. If the balance factor becomes 0 then the height of the subtree has decreased by one and the retracing needs to continue. If the balance factor becomes −2 or +2 then the subtree is unbalanced and needs to be rotated to fix it. If the rotation leaves the subtree's balance factor at 0 then the retracing towards the root must continue since the height of this subtree has decreased by one. This is in contrast to an insertion where a rotation resulting in a balance factor of 0 indicated that the subtree's height has remained unchanged.
The time required is O(log n) for lookup, plus a maximum of O(log n) rotations on the way back to the root, so the operation can be completed in O(log n) time.