AVL%20trees

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! )

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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 were 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)

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BST Delete breaks an AVL tree
7
7
Delete(9)
4
9
4
3
3
h(left) gt h(right)1 so NOT an AVL tree!
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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) lt -1 or bf(x) gt 1 then tree is NOT AVL

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Same example with bf(x), not h(x)
7
7
-2
-1
Delete(9)
4
9
4
-1
-1
0
3
3
0
0
bf lt -1 so NOT an AVL tree!
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What else can BST Delete break?

• Balance factors of ancestors

7
7
0
-1
-1
Delete(3)
9
4
9
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  the balance factors of ancestors to be outdated!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 lt -1 or gt 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  outdated also!

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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 subtrees 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
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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
h2
a
a
(deleted)
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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
2
z
T1
-1
h
a
T2
T3
h
h1
h1
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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
T3
T2
T1
h1
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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
T3
T2
T1
h
h1
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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
z
z
T1
T1
-1
-1
y
y
y
a
T3
T3
0
a
a
-1
1
T21
T22
T21
T22
T21
T22
h-1
h
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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
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Delete subcases for bf(x)2, bf(z)-1

• Case bf(y)0 double right-left rotation!

x
y
2
0
z
x
z
T1
-1
0
0
h
T1
T22
T21
T3
y
T3
0
h
T21
T22
h
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Delete subcases for bf(x)2, bf(z)-1

• Case bf(y)-1 double right-left rotation!

x
y
2
0
z
x
z
T1
-1
1
0
h
T1
T22
T21
T3
y
T3
-1
h
T21
T22
h-1
h
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Delete subcases for bf(x)2, bf(z)-1

• Case bf(y)1 double right-left rotation!

x
y
2
0
z
x
z
T1
-1
0
-1
h
T1
T22
T21
T3
y
T3
1
h
T22
T21
h-1
h
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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.

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Interactive AVL Deletes

• Interactive web applet