1 / 21

AVL trees

AVL trees. Today. AVL delete and subsequent rotations Testing your knowledge with interactive demos!. AVL tree. Is a binary search tree Has an additional height constraint : For each node x in the tree, Height(x.left) differs from Height(x.right) by at most 1 I promise:

beryl
Download Presentation

AVL trees

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. AVL trees

  2. Today • AVL delete and subsequent rotations • Testing your knowledge with interactive demos!

  3. AVL tree • Is a binary search tree • Has an additional height constraint: • For each node x in the tree, Height(x.left) differs from Height(x.right) by at most 1 • I promise: • If you satisfy the height constraint, then the height of the tree is O(lg n). • (Proof is easy, but no time! =])

  4. AVL tree • To be an AVL tree, must always: • (1) Be a binary search tree • (2) Satisfy the height constraint • Suppose we start with an AVL tree, then delete as if we’re in a regular BST. • Will the tree be an AVL tree after the delete? • (1) It will still be a BST… • (2) Will it satisfy the height constraint? • (Not covering insert, since you already did in class)

  5. BST Delete breaks an AVL tree 7 7 Delete(9) 9 4 4 3 3 h(left) > h(right)+1 so NOT an AVL tree!

  6. Replacing the height constraint with balance factors • Instead of thinking about the heights of nodes, it is helpful to think in terms of balance factors • The balance factor bf(x) = h(x.right) – h(x.left) • bf(x) values -1, 0, and 1 are allowed • If bf(x) < -1 or bf(x) > 1 then tree is NOT AVL

  7. Same example with bf(x), not h(x) 7 7 -2 -1 Delete(9) 9 4 4 -1 -1 0 3 3 0 0 bf < -1 soNOT an AVL tree!

  8. What else can BST Delete break? • Balance factors of ancestors… 7 7 0 -1 -1 Delete(3) 9 9 4 4 0 -1 -1 0 0 3 0

  9. Need a new Delete algorithm • Goal: if tree is AVL before Delete, then tree is AVL after Delete. • Step 1: do BST delete. • This maintains the BST property, but cancause the balance factors of ancestorsto be outdated! • Step 2: fix the height constraint and update balance factors. • Update any invalid balance factors affected by delete. • After updating them, they can be < -1 or > 1. • Do rotations to fix any balance factors that are too small or large while maintaining the BST property. • Rotations can cause balance factors to be outdated also!

  10. Bad balance factors • Start with an AVL tree, then do a BST Delete. • What bad values can bf(x) take on? • Delete can reduce a subtree’s height by 1. • So, it might increase or decrease h(x.right) – h(x.left) by 1. • So, bf(x) might increase or decrease by 1. • This means: • if bf(x) = 1 before Delete, it might become 2. BAD. • If bf(x) = -1 before Delete, it might become -2. BAD. • If bf(x) = 0 before Delete, then it is still -1, 0 or 1. OK. 2 cases

  11. Problematic cases for Delete(a) • bf(x) = -2 is just symmetric to bf(x) = 2. • So, we just look at bf(x) = 2. x x -2 2 h h+2 a a (deleted)

  12. Delete(a): 3 subcases for bf(x)=2 • Since tree was AVL before, bf(z) = -1, 0 or 1 Case bf(z) = -1 Case bf(z) = 0 Case bf(z) = 1 x x x 2 2 2 z z z T1 T1 T1 -1 0 1 h a a a T2 T3 T2 T2 T3 T3 h h+1 h+1

  13. Fixing case bf(x) = 2, bf(z) = 0 • We do a single left rotation • Preserves the BST property, and fixes bf(x) = 2 x z 2 -1 z x T3 T1 0 1 h T2 T2 T3 T1 h+1

  14. Fixing case bf(x) = 2, bf(z) = 1 • We do a single left rotation (same as last case) • Preserves the BST property, and fixes bf(x) = 2 x z 2 0 z x T3 T1 1 0 h T2 T2 T3 T1 h h+1

  15. Delete(a): bf(x)=2, bf(z)=-1 subcases Case bf(z) = -1: we have 3 subcases. (More details) Case bf(y) = 0 Case bf(y) = -1 Case bf(y) = 1 x x 2 2 x 2 z z T1 T1 -1 -1 z T1 -1 h y y y a T3 T3 0 a a -1 1 T3 h T21 T22 T21 T22 T21 T22 h-1 h

  16. Double right-left rotation • All three subcases of bf(x)=2, bf(z)=-1 simply perform a double right-left rotation. x x y z y z x y z

  17. Delete subcases for bf(x)=2, bf(z)=-1 • Case bf(y)=0: double right-left rotation! x y 2 0 z z x T1 0 -1 0 h T22 T1 T21 y T3 T3 0 h T21 T22 h

  18. Delete subcases for bf(x)=2, bf(z)=-1 • Case bf(y)=-1: double right-left rotation! x y 2 0 z z x T1 1 -1 0 h T22 T1 T21 y T3 T3 -1 h T21 T22 h-1 h

  19. Delete subcases for bf(x)=2, bf(z)=-1 • Case bf(y)=1: double right-left rotation! x y 2 0 z z x T1 0 -1 -1 h T22 T1 T21 y T3 T3 1 h T22 T21 h-1 h

  20. Recursively fixing balance factors • Idea: start at the node we deleted, fix a problem, then recurse up the tree to the root. • At each node x, we update the balance factor: bf(x) := h(bf.right) - h(bf.left). • If bf(x) = -2 or +2, we perform a rotation. • Then, we update the balance factors of every node that was changed by the rotation. • Finally, we recurse one node higher up.

  21. Interactive AVL Deletes • Interactive web applet

More Related