Java [1] Performing a check of constraints on node values for each sub-tree, just the way we discussed in the Lecture this week. Please remember what we dis[1]cussed in the Lecture - that - for a Binary Tree to qualify as a Binary Search Tree, we must perform the node values check of-course but also must not forget the value checks at the sub-tree level. [2] Performing the BST check by doing an In-Order Traversal of the Binary Tree as discussed in the Lecture. Since we know that an in-order traversal of a BST results in nodes being processed in sorted order, as soon as there is a violation of sorted order, we would know that the tree provided is not a BST.
Java [1] Performing a check of constraints on node values for each sub-tree, just the way we discussed in the Lecture this week. Please remember what we dis[1]cussed in the Lecture - that - for a Binary Tree to qualify as a Binary Search Tree, we must perform the node values check of-course but also must not forget the value checks at the sub-tree level. [2] Performing the BST check by doing an In-Order Traversal of the Binary Tree as discussed in the Lecture. Since we know that an in-order traversal of a BST results in nodes being processed in sorted order, as soon as there is a violation of sorted order, we would know that the tree provided is not a BST.
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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[1] Performing a check of constraints on node values for each sub-tree, just the way we discussed in the Lecture this week. Please remember what we dis[1]cussed in the Lecture - that - for a Binary Tree to qualify as a Binary Search Tree, we must perform the node values check of-course but also must not forget the value checks at the sub-tree level.
[2] Performing the BST check by doing an In-Order Traversal of the Binary Tree as discussed in the Lecture. Since we know that an in-order traversal of a BST results in nodes being processed in sorted order, as soon as there is a violation of sorted order, we would know that the tree provided is not a BST.
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