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Figure 12.15 Binary trees
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12.1 The disadvantage of a binary search tree is that it may become very unbalanced, in which case searching degenerates into an O(n) algorithm. The advantage is the efficiency that a binary tree provides for insertions and deletions. The advantage of an AVL tree is that it is always balanced, guaranteeing the O(lgn) speed of the binary search algorithm. The disadvantages the complex rotations used by the insertion and removal algorithms needed to maintain the tree s balance.
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12.1 To insert a new record with key 16 into the tree shown in Figure 12.16, the initial search would lead to the first leaf node. Since that is a five-way search tree, that first leaf node has overflowed, causing it to be split into two leaf nodes and moving its middle key 19 up to its parent node, as shown in Figure 12.17. But now that parent node has overflowed. So it also gets split, moving its middle key up to its parent node, as shown in Figure 12.18. Two other ordering of the seven keys in Example 12.5 on page 235 that will produce the same BST: a. 44, 22, 33, 77, 55, 99, 88 b. 44, 22, 77, 33, 55, 99, 88 An array of objects could be sorted by inserting their objects into a binary search tree and then using an inorder traversal to copy them back into the array. The BST property guarantees that the inorder traversal will visit the elements in order. If an AVL tree is used, then each insertion runs in O(lgn) time, so building the tree with n elements will require O(n lgn) time. The subsequent inorder traversal also has O(n lgn) complexity, so the entire algorithm sorts the array in O(n lgn) time. All except a are binary search trees.
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Figure 12.16 Inserting the key 16 in a five-way search tree
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Figure 12.17 Inserting the key 16 in a five-way search tree
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Figure 12.18 Inserting the key 16 in a five-way search tree
private AVLTree rotateRight() { AVLTree x = this, y = left, z = y.left; x.left = z; y.left = x; int xb = x.balance; int yb = y.balance;
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if (yb > 1) { ++x.balance; y.balance = ( xb<0 yb+1 : xb+yb+2 ); } else if (yb > xb) { x.balance += yb-1; ++y.balance; } else { y.balance = xb+2; } return y; }
Theorem. Every subtree of a binary search tree is a binary search tree. Proof: Let T be a binary search tree, and let S be a subtree of T. Let x be any element in S, and let L and R be the left and right subtrees of x in S. Then, since S is a subtree of T, x is also an element of T, and L and R are the left and right subtrees of x in T. Therefore, y x z for every y L and every z R because T has the BST property. Thus, S also has the BST property.
Theorem. Every subtree of an AVL tree is an AVL tree. Proof: The proof that every subtree of a binary search tree is a binary search tree is given in Problem 12.6. If a S is a subtree of an AVL tree T, then every node is S is also in T. Therefore, the balance number at every node in S is 1, 0, or 1.
The solution is shown in Figure 12.19.
Figure 12.19 AVL tree insertions
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Figure 12.19 (continued) AVL tree insertions
Heaps and Priority Queues
HEAPS A heap is a complete binary tree whose elements have keys that satisfy the following heap property: the keys along any path from root to leaf are descending (i.e., nonincreasing). EXAMPLE 13.1 A Heap
Figure 13.1 shows a heap. Note that the keys along each of its root-to-leaf paths are descending: 77 66 44 22; 77 66 44 41; 77 66 60 58; 77 66 60 25; 77 55 33 29; 77 55 55.
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