Binary Balanced Search Trees
CS-2851 Dr. Mark L. Hornick
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For any balanced binary search tree, searching, inserting and deleting have worstTime(n) that is logarithmic in n
But insertions and deletions can result in a tree that is highly unbalanced
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O(log n) performance drifts toward O(n)
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A BST is balanced only if if its height is logarithmic in n
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N=8, h=log2(8) = 3
CS-2851 Dr. Mark L. Hornick
N=5, h > log2(5) = 1.6
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An AVL tree is a balanced BST that either is empty or in which
eights between left and right subtrees differs by only 1
Unbalanced vs. balanced:
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he left and right subtrees are also AVL trees.
CS-2851 Dr. Mark L. Hornick
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Another (balanced) AVL Tree
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CS-2851 Dr. Mark L. Hornick
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More AVL Tree examples
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CS-2851 Dr. Mark L. Hornick
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Non-AVL Tree examples
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CS-2851 Dr. Mark L. Hornick 6
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Balance can be maintained through rotations
Rotation: an adjustment to the tree, around an element, that maintains the required ordering of elements
Right rotation around element 90
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CS-2851 Dr. Mark L. Hornick
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For any right rotation around element x, the right subtree of xs left child becomes x y the left subtree of x. y x
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Rotate right around 100: 100 ? 80 120
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CS-2851 Dr. Mark L. Hornick
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For any right rotation around element x, the right subtree of xs left child becomes x y the left subtree of x. y x
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Here is a right rotation around 100: 100 80
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CS-2851 Dr. Mark L. Hornick
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Rotate about right 20
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CS-2851 Dr. Mark L. Hornick 10
Right rotation about 20
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CS-2851 Dr. Mark L. Hornick
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For a right rotation around x
x.left = lc.right; lc.right = x; where x is a reference to an Entry object, and lc is a reference to the left child of x
Here is a right rotation around 100: 100 80
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50 Notice that 90 is now in the right subtree.
CS-2851 Dr. Mark L. Hornick
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In a rotation around x, the only restructuring is to the subtree rooted at x
right rotation around 100:
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CS-2851 Dr. Mark L. Hornick
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But the parent of x has to be adjusted (text p 421):
private void rotateRight(Entry<E> x) { Entry<E> lc = x.left; x.left = lc.right; // From previous slide if (lc.right != null) lc.right.parent = x; lc.parent = x.parent; if (x.parent == null) // x is root root = lc; else if (x.parent.right == x) // x is right child x.parent.right = lc; else // x is left child x.p...