ssrs 2d barcode Figure 11.28 A binary tree in Java

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Figure 11.28 A binary tree
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11.23 Show that, for all n 8, the function f(n) derived in Problem 11.22 produces the same sequence as the following explicit formula 2n n 2n ! - 2n 2n 1 2n 2 2n + 3 2n + 2 f n = ----------- = ----------------------- --------------------------------------------------------------------------------------------------n+1 n! n + 1 ! n n 1 n 2 n 3 2 1 For example, 8 4 8 7 8 7 6 8! f 4 = ------- = --------- = ------------------------------- = --------------- = 14 4 4 3 2 1 4!5! 5 11.24 Prove Corollary 11.3 on page 203. 11.25 Prove Theorem 11.2 on page 205.
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BINARY TREES
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11.26 Draw the forest that is represented by the binary tree shown in Figure 11.29. 11.27 Derive an explicit formula for the number f(h) of complete binary trees of height h. 11.28 Derive an explicit formula for the number f(h) of full binary trees of height h. 11.29 Implement the each of the following methods for the BinaryTree class: a. public int leaves(); b. c. d. e.
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Figure 11.29 A binary tree
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// returns the number of leaves in this tree public int height(); // returns the height of this tree public int level(Object object); // returns -1 if the given object is not in this tree; // otherwise, returns its level in this tree; public void reflect(); // swaps the children of each node in this tree public void defoliate(); // removes all the leaves from this tree
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11.1 11.2 11.3 11.4 11.5 11.6 11.7 The full binary tree of height 3 has l = 23 = 8 leaves. The full binary tree of height 3 has m = 23 1 = 7 internal nodes. The full binary tree of height 3 has n = 23+1 1 = 24 1 = 16 1 = 15 nodes. The full binary tree of height 9 has l = 29 = 512 leaves. The full binary tree of height 9 has m = 29 1 = 512 1 = 511 internal nodes. The full binary tree of height 9 has n = 29+1 1 = 210 1 = 1024 1 = 1023 nodes.
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h n 1. Thus in a binary tree with 100 nodes lg By Corollary 11.3, in any binary tree: lg n h 100 1 = 99. Since lg 100 = (log100)/(log2) = 6.6 = 6, it follows that the height must be between 6 and 99, inclusive: 6 h 99.
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The inorder traversal algorithm for binary trees recursively visits the root in between traversing the left and right subtrees. This presumes the existence of exactly two (possibly empty) subtrees at every (nonempty) node. In general trees, a node may have any number of subtrees, so there is no simple algorithmic way to generalize the inorder traversal. a. b. c. d. True True True False
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11.1 The equivalent trees are shown in Figure 11.30.
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Figure 11.30 Binary trees
The order of visitation in the binary tree traversal: a. Level order: A, B, C, D, E, F, G, H, I, J, K b. Preorder: A, B, D, E, H, I, C, F, J, G, K c. Inorder: D, B, H, E, I, A, F, J, C, G, K d. Postorder: D, H, I, E, B, J, F, K, G, C, A The order of visitation in the binary tree traversal: a. Level order traversal: A, B, C, D, E, F, H, I, J, M b. Preorder traversal: A, B, D, H, I, E, J, C, F, M c. Inorder traversal: H, D, I, B, J, E, A, F, M, C d. Postorder traversal: H, I, D, J, E, B, M, F, C, A The order of visitation in the binary tree traversal: a. Level order traversal: A, B, C, D, E, F, G, H, J, K, L, M, N, O b. Preorder traversal: A, B, D, G, M, H, C, E, J, N, F, K, O, L c. Inorder traversal: G, M, D, H, B, A, N, J, E, C, K, O, F, L d. Postorder traversal: M, G, H, D, B, N, J, E, O, K, L, F, C, A The natural mapping of the specified binary tree is shown in Figure 11.31.
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