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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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.
Transcribed Image Text:/* Class to represent Tree node */
class Node {
int data;
Node left, right;
public Node(int item)
data = item;
left = null;
right - null;
The root element of the Binary Tree is given to you. Below is an illustrated sam-
ple of Binary Tree nodes for your reference, which in-fact is the same example
we discussed in the lecture.
tree.root - new Node (4);
tree.root.left = new Node (2);
tree.root.right = new Node (6);
tree.root.left.left = nev Node(1);
tree.root.left.right = new Node (3);
tree.root.right.left = new Node (5);
tree.root.right.right = new Node (7);
Your code will need to return a boolean : True or False.
When you follow the validation process specified - the complexity of the solution
will be as below.
Time Complexity: 0(n)
Space Complexity: 0(n)
The linear space complexity would come from the recursion (AKA "recursion
stack") you employ to validate the Tree.
Submissions that don't meet the linear Time and Space complexities will only
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