Java [1] Delete the root node value of the BST and replace the root value with the appropriate value of the existing BST.

Computer Networking: A Top-Down Approach (7th Edition)
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Author:James Kurose, Keith Ross
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Chapter1: Computer Networks And The Internet
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Java

[1] Delete the root node value of the BST and replace the root value with

the appropriate value of the existing BST.

 

[2] Perform the BST status check by doing an In-Order Traversal of the

BST such that even after deletion the BST is maintained. Also, a short paragraph for the time and space complexity for your code

/* 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 Search Tree is given to you. Below is an illus-
trated sample of Binary Search 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 = new 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 delete the root node value provided and replace the root value
with appropriate value such that the BST property is maintaned which would
be known by performing the In-Order Traversal of the new BST. Obviously, the
In-order traversal of the resulting BST would need to be ascending order
When you follow the deletion process and the BST traversal specified - the com-
plexity of the solution will be as below.
Time Complexity: 0(n)
Space Complexity: 0(n)
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 Search Tree is given to you. Below is an illus- trated sample of Binary Search 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 = new 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 delete the root node value provided and replace the root value with appropriate value such that the BST property is maintaned which would be known by performing the In-Order Traversal of the new BST. Obviously, the In-order traversal of the resulting BST would need to be ascending order When you follow the deletion process and the BST traversal specified - the com- plexity of the solution will be as below. Time Complexity: 0(n) Space Complexity: 0(n)
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