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public AVLTree attach(Object object) { if (root == null) { // tree is empty root = object; return this; } if (comp.compare(object,root) < 0) { // insert into left subtree if (left == null) { left = new AVLTree(object,comp); ++size; --balance; } else { int lb = left.balance; left = left.attach(object); if (left.balance != lb && left.balance != 0) { --balance; } } if (balance < -1) { if (left.balance > 0) { left = left.rotateLeft(); } return rotateRight(); } } else { // insert into right subtree if (right == null) { right = new AVLTree(object,comp); ++size; ++balance; } else { int rb = right.balance; right = right.attach(object); if (right.balance != rb && right.balance != 0) { ++balance; } } if (balance > 1) { if (right.balance < 0) { right = right.rotateRight(); } return rotateLeft(); } } return this; } private AVLTree rotateRight() // see Problem 12.5 on page 240
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private AVLTree rotateLeft() { AVLTree x = this, y = right, z = y.left; x.right = z; y.left = x; int xb = x.balance, yb = y.balance; if (yb < 1) { --x.balance; y.balance = ( xb>0 yb-1 : xb+yb-2 );
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} else if (yb < xb) { x.balance -= yb+1; --y.balance; } else { y.balance = xb-2; } return y;
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EXAMPLE 12.8 Building an AVL Tree
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Insertions of G, M, T, D, and P into an empty AVL tree are shown in Figure 12.13.
Figure 12.13 Inserting into an AVL tree
The first rotation occurs with the insertion of T. That increases the balance at the root to 2, which violates the AVL constraint. The left rotation about the root x makes M become the parent of its prior parent G. The next rotation occurs after E in inserted. The right rotation at its parent D straightens out the dogleg G D E but leaves the balance at G at 2. That requires a second rotation in the opposite direction. Double rotations like this are required when the imbalance is at the top of a dogleg. Note how efficient the rotations are. By making only local changes to references and balance numbers, they restore the tree to nearly perfect balance. Figure 12.14 on page 240 shows a later insertion into the same AVL tree, inserting W after U, V, and Z have been inserted. This illustrates a double rotation where a nontrivial subtree gets shifted. The subtree containing U is shifted from parent V to parent T. Note that the BST property is maintained.
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Figure 12.14 AVL tree rotations
Although a bit complicated, the insertion algorithm for AVL trees is very efficient. The rotations that keep it balanced make only local changes to a few references and balance numbers. Review Questions
12.1 12.2 What are the advantages and disadvantages of using a binary search tree What are the advantages and disadvantages of using an AVL tree
Problems
12.1 12.2 12.3 12.4 12.5 12.6 12.7 12.8 Describe what happens in the five-way tree shown in Example 12.1 on page 230 when a new record with key 16 is inserted. Find two other orderings of the seven keys in Example 12.5 on page 235 that will produce the same binary search tree. Describe a method for sorting arrays of objects using a binary search tree. Then determine the complexity of the algorithm. Determine which of the binary trees shown in Figure 12.15 on page 241 is a binary search tree. Write the rotateRight() method for the AVLTree class. Prove that every subtree of a binary search tree is also a binary search tree. Prove that every subtree of an AVL tree is also an AVL tree. Here are the U.S. Postal Service abbreviations of the first 10 states, in the order that they ratified the U.S. Constitution: DE, PA, NJ, GA, CT, MA, MD, SC, NH, VA. Show the AVL tree after the insertion of each of these strings.
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