Understanding AVL Tree Rotations: Solving the Case RL

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In summary, we are at the case RL in an AVL-tree where $58$ is being inserted, and two rotations are required to balance the tree. This is because the first edge from the critical point to the new-inserted element is a left one and the second is a right one, indicating the case RL.
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evinda
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Hello! (Smile)We want to insert $58$ at the following AVL-tree and then we have to make rotations so that the tree is balanced. According to my notes, we are at the case [m]RL[/m] (The first edge leads to the right and the second to the left), where two rotations are required, one right around the next of the critical node and a left one, around the critical node.

View attachment 3879

But why are we at the case RL, although the first edge from the critical point to the new-inserted element is a left one and the second a right one? (Thinking)PS: The first node at the path from the node v (the new-inserted node) to the root, of which [m]balance=ReightHeight(v)-LeftHeight(v)[/m] was [m]-1[/m] or [m]1[/m](before the insertion) is called critical node.
 

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We are at the case RL because we are considering the path of edges from the node v (the new-inserted node) to the root. The first edge leading from the critical point to the new-inserted element is a left one and the second a right one, so this is the case RL.
 

Related to Understanding AVL Tree Rotations: Solving the Case RL

What is an AVL-tree?

An AVL-tree is a self-balancing binary search tree. It is named after its inventors, Adelson-Velskii and Landis. It maintains a balance factor for each node, which is the difference between the heights of its left and right subtrees. This balance factor is always kept within a range of -1, 0, or 1, ensuring that the tree remains balanced.

How do you insert an element in an AVL-tree?

To insert an element in an AVL-tree, you start at the root and traverse the tree according to the rules of a binary search tree until you find the appropriate position for the new element. Then, you add the element as a new leaf and update the balance factor for all the nodes on the path from the inserted node to the root. If the balance factor of any node becomes greater than 1 or less than -1, the tree needs to be rebalanced.

What is the time complexity of inserting an element in an AVL-tree?

The time complexity of inserting an element in an AVL-tree is O(log n), where n is the number of nodes in the tree. This is because the tree is self-balancing, so the height of the tree remains logarithmic.

Why is it important to maintain balance in an AVL-tree?

Maintaining balance in an AVL-tree is important because it ensures efficient searching, insertion, and deletion operations. If the tree is unbalanced, these operations can take longer as the tree becomes deeper, resulting in a time complexity of O(n) instead of O(log n). Additionally, an unbalanced tree can lead to skewed data distribution, which can affect the performance of the tree.

Can an AVL-tree have duplicate elements?

No, an AVL-tree cannot have duplicate elements. This is because AVL-trees are binary search trees, and by definition, binary search trees do not allow duplicate elements. If a duplicate element is inserted, it will either be rejected or will replace the existing element, depending on the implementation.

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