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# Construct Diamond Tree

The Construct Diamond Tree problem involves creating a special binary tree known as a Diamond Tree. A Diamond Tree is a unique binary tree where each node has two children, forming a diamond pattern. The goal is to construct this tree and then print its elements in a specific level order.

## Problem Statement

Given an integer `k`, the task is to construct a Diamond Tree of `k` levels and then print its elements in a specific level order.

## Example

Let's consider `k = 4`:

The constructed Diamond Tree would look like this: The printed level order of the Diamond Tree would be:

``````1
2 3
4 5 6 7
8 9 10 11 12 13 14 15
17 21 25 29
38 54
92``````

## Idea to Solve

The problem can be solved by constructing the Diamond Tree using a queue-based approach. The main idea is to iteratively add nodes to the queue, constructing the tree level by level. Then, print the elements in a level-wise manner.

## Algorithm

1. Initialize a queue to hold tree nodes.
2. Create the root node with the value 1.
3. Enqueue the root node into the queue.
4. Loop while `k` is greater than 1 (to construct the Diamond Tree):
• Get the current size of the queue (`n`).
• Loop `n` times:
• Dequeue a node from the queue.
• Create left and right children nodes with values `num + 1` and `num + 2`.
• Enqueue the left and right children nodes into the queue.
• Increment `num` by 2.
• Decrement `k` by 1.
5. Loop while the queue size is greater than 1 (to construct the bottom part of the Diamond Tree):
• Dequeue a node (first node).
• Dequeue another node (second node).
• Create a new node with the sum of the keys of the two dequeued nodes.
• Connect the new node to the right side of the first dequeued node.
• Connect the new node to the left side of the second dequeued node.
• Enqueue the new node into the queue.
6. Print the elements of the Diamond Tree in a level order:
• Enqueue the root node into the queue.
• While the queue is not empty:
• Get the current size of the queue (`n`).
• Loop `n` times:
• Dequeue a node from the queue.
• Enqueue its left and right children if they exist and are not the same as the previously dequeued node.
• Print the key of the dequeued node.
• Move to the next line to signify a change in levels.

## Time Complexity

The time complexity of constructing the Diamond Tree and printing it in level order is O(N), where N is the total number of nodes in the Diamond Tree. This is because each node is enqueued and dequeued exactly once during the construction and traversal processes.

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