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 queuebased approach. The main idea is to iteratively add nodes to the queue, constructing the tree level by level. Then, print the elements in a levelwise manner.
Algorithm
 Initialize a queue to hold tree nodes.
 Create the root node with the value 1.
 Enqueue the root node into the queue.
 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
andnum + 2
.  Enqueue the left and right children nodes into the queue.
 Increment
num
by 2.
 Decrement
k
by 1.
 Get the current size of the queue (
 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.
 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.
 Get the current size of the queue (
 Move to the next line to signify a change in levels.
Code Solution

1) Diamond tree implementation program in java
2) Diamond tree implementation in c++
3) Diamond tree implementation in c
4) Diamond tree implementation in c#
5) Diamond tree implementation in php
6) Diamond tree implementation in python
7) Diamond tree implementation in ruby
8) Diamond tree implementation in scala
9) Diamond tree implementation in swift
10) Diamond tree implementation in kotlin
11) Diamond tree implementation in node js
12) Diamond tree implementation in golang
13) Diamond tree implementation in vb.net
14) Diamond tree implementation in typescript
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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