dynamic operations on binary search tree mostly requires $O(h)$ time

the minimum $h$ of a tree of $n$ nodes is $logn$

dynamic operations such as insertion may lead to the height of a tree way larger than $logn$, which then increase operation time

• the extreme case is that new items are being added in ascending order, the tree will be one long branch off to the right, with height of $n$

this lecture shows how to balance the height of binary tree

how to maintain height $h=O(logn)$ where n is the number of nodes in the tree?

a binrayr tree that maintains $O(logn)$ height under dynamic operations is called balanced.

• many algorithms

• such as AVL tree

need reduce height of tree without changing its traversal order so that we represent the same sequence of items

Rotations!

A rotation relinks $O(1)$ ppinters to modify tree structure and maintains traversal order

AVL Trees maintain height-balance

• A node is height-balanced if heights of its left and right subtrees differ by at most 1

• let skew of a node be the height of its right subtree minus that of its left subtree

• then a node is height-balanced if its skew is -1, 0, or 1

how to tell wether a node is height-balanced? compute the height of subtrees!!

algorithm

• recusively compute the height of left subtree and right subtree of node <X>

• add 1 to the max of the two heights

• run in $O(n)$ time since we recurse on every node

can we improve to $O(1)$? -> tree augumentation: augument each node with the height of its subtree !!

the height of <X> can be computed in $O(1)$ time from the heights of its children

• look up the stored height of left and right child in $O(1)$

• add 1 to the max of the two

during dynamic operaiton, the height and other augumentation of each node must be maintained as the tree changes shape

• update relinked nodes in a rotation operation in $O(1)$ time

• update all ancestors of an inserted or deleted node in $O(h)$ time by walking up the tree.