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AVL trees

AVL trees. AVL Trees. We have seen that all operations depend on the depth of the tree. We don’t want trees with nodes which have large height This can be attained if both subtrees of each node have roughly the same height.

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AVL trees

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  1. AVL trees

  2. AVL Trees We have seen that all operations depend on the depth of the tree. We don’t want trees with nodes which have large height This can be attained if both subtrees of each node have roughly the same height. • AVL tree is a binary search tree where the height of the two subtrees of a node differs by at most one • Height of a null tree is -1

  3. 5 3 8 4 10 1 5 Not AVL Tree 3 4 1 AVL Tree

  4. Section 10.4 KR • Suppose an AVL tree of height h contains contains at most S(h) nodes: • S(h) = L(h) + R(h) + 1 • L(h) is the number of nodes in left subtree • R(h) is the number of nodes in right subtree You have larger number of nodes if there is larger imbalance between the subtrees This happens if one subtree has height h, another h-2 Thus, S(h) = S(h) + S(h-2) + 1

  5. Operations in AVL Tree O(log N) Searching, Complexity? O(log N) FindMin, Complexity? Deletion? Insertion?

  6. Insertion Search for the element If it is not there, insert it in its place. Any problem? Insertion may imbalance the tree. Heights of two children of a node may differ by 2 after an insertion. Tree Rotations used to restore the balance.

  7. If an insertion cause an imbalance, which nodes can be affected? Nodes on the path of the inserted node. • Let U be the node nearest to the inserted one which has an imbalance. • insertion in the left subtree of the left child of U • insertion in the right subtree of the left child of U • insertion in the left subtree of the right child of U • insertion in the right subtree of the right child of U

  8. V U X Z Y Insertion in left child of left subtree Single Rotation U V Z Y X After Rotation Before Rotation

  9. U Z V 3 1 Y 4 4 8 1 0.8 X 3 5 5 8 3 4 1 0.8 5 8 AVL Tree Insert 0.8 After Rotation

  10. W V V D D W U A B C C B Double Rotation • Suppose, imbalance is due to an insertion in the left subtree of right child • Single Rotation does not work! U Before Rotation After Rotation A

  11. U D V 3 A W 8 4 3.5 1 4 B 3.5 5 5 8 3 3 4 1 0.8 5 8 AVL Tree Insert 3.5 After Rotation

  12. 2 2 3 2 3 4 4 1 1 1 2 2 3 3 Fig 1 5 3 3 Fig 4 Fig 2 Fig 3 1 Fig 5 Fig 6 Extended Example Insert 3,2,1,4,5,6,7, 16,15,14

  13. 2 4 4 2 4 Fig 8 Fig 7 5 6 5 7 6 3 3 7 3 2 1 2 2 1 5 4 5 6 4 6 Fig 10 Fig 9 5 Fig 11 3 1 3 1 1

  14. 4 4 4 7 15 7 16 3 16 3 3 16 2 2 2 Fig 12 Fig 13 15 6 6 6 5 7 Fig 14 5 5 1 1 1

  15. 4 4 6 14 15 16 15 14 3 3 16 2 2 Fig 15 Fig 16 7 6 5 7 5 1 1 Deletions can be done with similar rotations

  16. Rings a bell! Fibonacci numbers FN N S(h) h h is O(log N) Using this you can show that h = O(log N)

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