What is the RUN TIME COMPLEXITY of the following code:
.................................................................................
public void printLevelsRecursively(Node root) {
for (int i = 1; i <= heightOfTree(root); i++) {
System.out.print("Level " + i + " : ");
printSingleLevelRecursively(root, i);
System.out.print("\n");
}
}
public int heightOfTree(Node root) {
if (root != null)
return super.max(heightOfTree(root.l),
heightOfTree(root.r)) + 1;
return 0;
}
public void printSingleLevelRecursively(Node root, int
level) {
if (root == null)
return;
if (level == 1)
System.out.print(root.key + " ");
else if (level > 1) {
printSingleLevelRecursively(root.l, level - 1);
printSingleLevelRecursively(root.r, level - 1);
}
}
Height Balanced Tree – (BST -binary search
tree)
AVL tree is binary search tree with additional property that
difference between height of left sub-tree and right sub-tree of
any node can’t be more than 1. For example, BST shown in Figure 2
is not AVL as difference between left sub-tree and right sub-tree
of node 3 is 2. However, BST shown Figure
figure 1 figure 2. figure 3
3 is AVL tree.
What is the RUN TIME COMPLEXITY of the following code: ................................................................................. public void printLevelsRecursively(Node root) {...
Using the following implementation of Tree class Node { public int iData; // data item (key) public double dData; // data item public Node leftChild; // this node's left child public Node rightChild; // this node's right child public void displayNode() // display ourself { System.out.print('{'); System.out.print(iData); System.out.print(", "); System.out.print(dData); System.out.print("} "); } } // end class Node //------------------------------------------------------------------ import java.io.IOException; import java.util.Stack; public class Tree { private Node root; // first node of tree // ------------------------------------------------------------- public Tree() // constructor { root = null; }...
Question - modify the code below so that for a node, the value of every node of its right subtree is less the node, and the value of each node of its left subtree is greater than the node. - create such a binary tree in the Main method, and call the following method: InOrder(Node theRoot), PreOrder(Node theRoot), PostOrder(Node theRoot), FindMin(), FindMax(), Find(int key) using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.Threading.Tasks;...
Convert into pseudo-code for below code =============================== class Main { public static void main(String args[]) { Scanner s=new Scanner(System.in); ScoresCircularDoubleLL score=new ScoresCircularDoubleLL(); while(true) { System.out.println("1--->Enter a number\n-1--->exit"); System.out.print("Enter your choice:"); int choice=s.nextInt(); if(choice!=-1) { System.out.print("Enter the score:"); int number=s.nextInt(); GameEntry entry=new GameEntry(number); ...
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Please provide the big oh notation for running time and space complexity for the following functions (available, into, out, size, and printAll): int SplayTreeInventory::available(string id){ Node* temp=stmap->findSplay(id); if(temp!=NULL) return temp->value; else return -1; } void SplayTreeInventory::into(string id, int amount) { Node* temp=stmap->findSplay(id); if(temp==NULL){ stmap->putSplay(id, amount); } else{ temp->key+=amount; } } void SplayTreeInventory::out(string id, int amount) { Node* temp=stmap->findSplay(id); if(temp==NULL){ } else{ if(temp->value<=amount) stmap->erase(id); else temp->value-=amount; } } int SplayTreeInventory::size() { return stmap->size(); } void SplayTreeInventory::printAll() { printAllHelper(this->stmap->root); }
Java Programming: The following is my code: public class KWSingleLinkedList<E> { public void setSize(int size) { this.size = size; } /** Reference to list head. */ private Node<E> head = null; /** The number of items in the list */ private int size = 0; /** Add an item to the front of the list. @param item The item to be added */ public void addFirst(E...
Q. write the algorithm for each function in this code: void insert(int x, node*&p) { //cheak if the pointer is pointing to null. if (p==NULL) { p = new node; p->key=x; p->left=NULL; p->right=NULL; } else { //Cheak if the element to be inserted is smaller than the root. if (x < p->key) { //call the function itself with new parameters. insert(x,p->left); } //cheak if the alement to be inserted...