AVL Trees concepts

In computer science, an AVL tree is a self-balancing binary search tree, and it is 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; therefore, it is also said to be height-balanced. 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.

The AVL tree is named after its two inventors, G.M. Adelson-Velskii and E.M. Landis, who published it in their 1962 paper "An algorithm for the organization of information."

The balance factor of a node is the height of its right subtree minus the height of its left subtree and a node with balance factor 1, 0, or -1 is considered balanced. A node with any other balance factor is considered unbalanced and requires rebalancing the tree. The balance factor is either stored directly at each node or computed from the heights of the subtrees.

AVL trees are often compared with red-black trees because they support the same set of operations and because red-black trees also take O(log n) time for the basic operations. AVL trees perform better than red-black trees for lookup-intensive applications.

Operations on AVL tree:

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

After inserting a node, it is necessary to check each of the node's ancestors for consistency with the rules of AVL. For each node checked, if the balance factor remains -1, 0, or 1 then no rotations are necessary. However, if the balance factor becomes 2 or -2 then the subtree rooted at this node is unbalanced. If insertions are performed serially, after each insertion, at most two tree rotations are need 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. Let R be the right child of P. Let L be the left child of P.

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. If the balance factor of R is +1, a left rotation is needed with P as the root. If the balance factor of R is -1, a double left rotation is needed. The first rotation is a right rotation with R as the root. The second is a left rotation with P as the root.(I think the last sentence is related to 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 then checked. If the balance factor of L is -1, a right rotation is needed with P as the root. If the balance factor of L is +1, a double right rotation is needed. The first rotation is a left rotation with L as the root. The second is a right rotation with P as the root. (I think the last sentence is related to left-right case).

Deletion

If the node is a leaf, remove it. If the node is not a leaf, replace it with either the largest in its left subtree (inorder predecessor) or the smallest in its right subtree (inorder successor), and remove that node. The node that was found as replacement has at most one subtree. After deletion, retrace the path back up the tree (parent of the replacement) to the root, adjusting the balance factors as needed.

In addition to the balancing described above for insertions, if the balance factor for the tree is 2 and that of the right subtree is 0, a left 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.

source - wikipedia

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