Introduction to Algorithms
3rd Edition
ISBN: 9780262033848
Author: Thomas H. Cormen, Ronald L. Rivest, Charles E. Leiserson, Clifford Stein
Publisher: MIT Press
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Chapter 15.5, Problem 1E
Program Plan Intro
To write the pseudo-code for the CONSTRUCT-OPTIMAL-BST( root ).
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Assuming that there are no duplicate keys in a binary search tree, develop
an algorithm to determine the post-order traversal of the tree based on its
pre-order traversal. For example, given the pre-order traversal: 5 214 3
12 10 9 15. The output is 1 3 4 2 9 10 15 12 5.
E.
D
2. In the binary search tree above, finding node E requires one comparison and finding
node A requires four comparisons. What is the expected number of comparisons
required to find a node chosen at random?
Consider the following traversal sequences:
Preorder traversal sequence: 74 72 45 75 67 82
Inorder traversal sequence: 45 72 75 74 82 67
Draw the corresponding binary tree.
Note: IMPORTANT: The penalty scheme for this question is 0, 50, 100%.
Answer: (penalty regime: 0, 50, 100 %)
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Chapter 15 Solutions
Introduction to Algorithms
Ch. 15.1 - Prob. 1ECh. 15.1 - Prob. 2ECh. 15.1 - Prob. 3ECh. 15.1 - Prob. 4ECh. 15.1 - Prob. 5ECh. 15.2 - Prob. 1ECh. 15.2 - Prob. 2ECh. 15.2 - Prob. 3ECh. 15.2 - Prob. 4ECh. 15.2 - Prob. 5E
Ch. 15.2 - Prob. 6ECh. 15.3 - Prob. 1ECh. 15.3 - Prob. 2ECh. 15.3 - Prob. 3ECh. 15.3 - Prob. 4ECh. 15.3 - Prob. 5ECh. 15.3 - Prob. 6ECh. 15.4 - Prob. 1ECh. 15.4 - Prob. 2ECh. 15.4 - Prob. 3ECh. 15.4 - Prob. 4ECh. 15.4 - Prob. 5ECh. 15.4 - Prob. 6ECh. 15.5 - Prob. 1ECh. 15.5 - Prob. 2ECh. 15.5 - Prob. 3ECh. 15.5 - Prob. 4ECh. 15 - Prob. 1PCh. 15 - Prob. 2PCh. 15 - Prob. 3PCh. 15 - Prob. 4PCh. 15 - Prob. 5PCh. 15 - Prob. 6PCh. 15 - Prob. 7PCh. 15 - Prob. 8PCh. 15 - Prob. 9PCh. 15 - Prob. 10PCh. 15 - Prob. 11PCh. 15 - Prob. 12P
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- The Tic-tac-toe game represents a 3x3 grid of two players X and Y. The winner finishes the game with 3 sequence X's (Player X) or O's (Player O). Use the Depth-first search to reach node G3 in the tree below and write the node visited in sequence time. Note: The answer is as a sequence of characters represents the state with comma separated. Example: A, GI, D, E, B, H, F, L, M, G2, G3, N o 0x Player X Player O oxx o 6 x Player X X X olox olxx oxx o ox x x x x xO olo x oxx Oxx M Answer:arrow_forwardTl and T2 are two very large binary trees, with Tl much bigger than T2. Create an algorithm to determine if T2 is a subtree of Tl.A tree T2 is a subtree of Tl if there exists a node n in Tl such that the subtree of n is identical to T2. That is, if you cut off the tree at node n, the two trees would be identical.arrow_forwardThe Tic-tac-toe game represents a 3×3 grid of two players X and Y. The winner finishes the game with 3 sequence X's (Player X) or O's (Player O). Use the Depth-first search to reach node G3 in the tree below and write the node visited in sequence time. Note: The answer is as a sequence of characters represents the state with comma separated. Example: A, GI, D, E, B, H, F, L, M, G2, G3, N o ox Player X o o x Player O Player X x X olox oxx oxx o 6x x x x x x O o ox o ox oxx oxx M Answer:arrow_forward
- 6. What is the time complexity of inserting a node into a balanced binary search tree with n nodes? Α. Ο (1) B. 0 (n) C. O (nlogn) D. O(n) E. O (logn)arrow_forwardLet T be a binary search tree with n nodes, in which n > 0. When T is linear, the search algorithm makes key comparisons, in the unsuccassful case.arrow_forwardThe BST remove algorithm traverses the tree from the root to find the node to remove. When the node being removed has 2 children, the node's successor is found and a recursive call is made. One node is visited per level, and in the worst-case scenario, the tree is traversed twice from the root to a leaf. A BST with N nodes has at least log2N levels and at most N levels. Therefore, the runtime complexity of removal is best case O(logN) and worst case O(N). Two pointers are used to traverse the tree during removal. When the node being removed has 2 children, a third pointer and a copy of one node's data are also used, and one recursive call is made. Thus, the space complexity of removal is always O(1)." I have to explain this clearly! and the advantages of the BST algorithimarrow_forward
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