# Utopian Tree

### Problem Statement :

```The Utopian Tree goes through 2 cycles of growth every year. Each spring, it doubles in height. Each summer, its height increases by 1 meter.

A Utopian Tree sapling with a height of 1 meter is planted at the onset of spring. How tall will the tree be after  growth cycles?

For example, if the number of growth cycles is n = 5, the calculations are as follows:

Period  Height
0          1
1          2
2          3
3          6
4          7
5          14

Function Description

Complete the utopianTree function in the editor below.

utopianTree has the following parameter(s):

int n: the number of growth cycles to simulate

Returns

int: the height of the tree after the given number of cycles

Input Format

The first line contains an integer, t, the number of test cases.
t subsequent lines each contain an integer, n, the number of cycles for that test case.

Constraints
1 <= t <= 10
0 <= n <= 60```

### Solution :

```                            ```Solution in C :

python 3  :

#!/usr/bin/env python

import sys

if __name__ == '__main__':

for _ in range(T):
height = 1

for i in range(N):
if i % 2 == 0:
height *= 2
else:
height += 1

print(height)

Java  :

import java.util.Scanner;
import java.util.Vector;

public class Solution {
public static void main(String[] args) {
Scanner stdin = new Scanner(System.in);
int numCases = stdin.nextInt();
Vector<Long> cache = new Vector<>();
for (int i = 0; i < numCases; i++) {
int numCycles = stdin.nextInt();
while(cache.size() <= numCycles) {
if (cache.size() % 2 == 0) {
} else {
}
}
System.out.println(cache.get(numCycles));
}
}
}

C ++  :

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

int main() {
int T;
cin >> T;
for(int t = 0; t<T; t++){
int N;
cin >> N;
int ht = 1;
for(int i=0;i<N;i++){
if(i%2==0) ht*=2;
else ht++;
}
cout << ht << endl;
}
return 0;
}

C  :

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {

/* Enter your code here. Read input from STDIN. Print output to STDOUT */
int t,i,n;
scanf("%d",&t);
while(t>0)
{
int h=1;
scanf("%d",&n);
for(i=1;i<=n;i++)
{
if(i%2==1)
h=2*h;
else
h=h+1;
}
printf("%d\n",h);
t=t-1;
}
return 0;
}```
```

## Tree: Inorder Traversal

In this challenge, you are required to implement inorder traversal of a tree. Complete the inorder function in your editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's inorder traversal as a single line of space-separated values. Input Format Our hidden tester code passes the root node of a binary tree to your \$inOrder* func

## Tree: Height of a Binary Tree

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## Tree : Top View

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## Tree: Level Order Traversal

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## Binary Search Tree : Insertion

You are given a pointer to the root of a binary search tree and values to be inserted into the tree. Insert the values into their appropriate position in the binary search tree and return the root of the updated binary tree. You just have to complete the function. Input Format You are given a function, Node * insert (Node * root ,int data) { } Constraints No. of nodes in the tree <

## Tree: Huffman Decoding

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