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c. The leaf nodes are C, E, G, J, L, N, O, P, W, Y, and Z; node C has no children; node F has depth 2; the nodes at level 3 are H, J, and K; the height of the tree is 9; the order of the tree is 3. d. The leaf nodes are G, H, K, L, N, O, P, Q, R, S, and T; the only child node C has is node E; node F has depth 3; the nodes at level 3 are F, G, H, and J; the height of the tree is 5; the order is 5. e. The leaf nodes are D, E, L, N, P, Q, R, S, and T; node C has no children; node F has depth 1; the nodes at level 3 are K, L, M, N, and O; the height of the tree is 4; the order of the tree is 5. 10.4 a. b. c. d. a. b. c. d. The ancestors of F are C and A The descendants of F are I, K, and L. The nodes in the subtree rooted at F are F, I, K, and L. The leaf nodes are D, H, J, K, and L. (35 1)/2 = 121 nodes (44 1)/3 = 85 nodes (105 1)/9 = 11,111 nodes (411 1)/3 = 1,398,101 nodes
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a. Level order: a, b, c, d, e, f, g, h, i, j, k, l, m, n, o. b. Preorder: a, b, e, i, j, c, f, k, g, h, l, m, n, o, d. c. Postorder: i, j, e, b, k, f, g, l, m, n, o, h, c, d, a. a. b. c. d. The level order and the preorder traversals always visit the root first. The postorder traversal always visits the left-most node first. The postorder traversal always visits the root last. The preorder traversal always visits the right-most node last.
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10.8 10.9
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The preorder traversal follows the pattern of reading by column from left to right. The preorder traversal is used in Problem 9.32 on page 184.
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10.1 Proof of Theorem 10.1 on page 187: If there were no path from a given node x to the root of the tree, then the definition of tree would be violated, because to be an element of the tree, x must be the root of some subtree. If there were more than one path from x back to the root, then x would be an element of more than one distinct subtree. That also violates the definition of tree, which requires subtrees to be disjoint. Proof of Theorem 10.2 on page 188: If the tree is empty, then its height is h = 1 and the number of nodes n = 0. In that case, the formula is correct: n = (d (h)+1 1)/(d 1) = (d ( 1)+1 1)/(d 1) = (d 0 1)/(d 1) = (1 1)/(d 1) = 0. If the tree is a singleton, then its height is h = 0 and the number of nodes n = 1. In that case, the formula is again correct: n = (d (h)+1 1)/(d 1) = (d (0)+1 1)/(d 1) = (d 1)/(d 1) = 1. Now assume that the formula is correct for any full tree of height h 1, where h 0. Let T be the full tree of height h. Then by definition, T consists of a root node and a set of d subtrees. And since T is full, each of its d subtrees has height h 1. Therefore, by the inductive hypothesis, the number of nodes in each subtree is nS = (d (h 1)+1 1)/(d 1) = (d h 1)/(d 1). Thus, the total number of nodes in T is n = 1 + d nS dh 1 = 1 + d ------------d 1 d 1 dh + 1 d = ----------- + -------------------d 1 d 1
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