Construct a complete binary tree from given array in level order fashion
Given an array of elements, our task is to construct a complete binary tree from this array in a level order fashion. That is, elements from the left in the array will be filled in the tree level-wise starting from level 0.
Examples:
Input : arr[] = {1, 2, 3, 4, 5, 6}
Output : Root of the following tree
1
/ \
2 3
/ \ /
4 5 6
Input: arr[] = {1, 2, 3, 4, 5, 6, 6, 6, 6, 6}
Output: Root of the following tree
1
/ \
2 3
/ \ / \
4 5 6 6
/ \ /
6 6 6
If we observe carefully we can see that if the parent node is at index i in the array then the left child of that node is at index (2*i + 1) and the right child is at index (2*i + 2) in the array.
Using this concept, we can easily insert the left and right nodes by choosing their parent node. We will insert the first element present in the array as the root node at level 0 in the tree and start traversing the array and for every node, we will insert both children left and right in the tree.
Below is the recursive program to do this:
C++
// CPP program to construct binary // tree from given array in level // order fashion Tree Node #include <bits/stdc++.h> using namespace std; /* A binary tree node has data, pointer to left child and a pointer to right child */struct Node { int data; Node* left, * right; }; /* Helper function that allocates a new node */Node* newNode(int data) { Node* node = (Node*)malloc(sizeof(Node)); node->data = data; node->left = node->right = NULL; return (node); } // Function to insert nodes in level order Node* insertLevelOrder(int arr[], int i, int n) { Node *root = nullptr; // Base case for recursion if (i < n) { root = newNode(arr[i]); // insert left child root->left = insertLevelOrder(arr, 2 * i + 1, n); // insert right child root->right = insertLevelOrder(arr, 2 * i + 2, n); } return root; } // Function to print tree nodes in // InOrder fashion void inOrder(Node* root) { if (root != NULL) { inOrder(root->left); cout << root->data <<" "; inOrder(root->right); } } // Driver program to test above function int main() { int arr[] = { 1, 2, 3, 4, 5, 6, 6, 6, 6 }; int n = sizeof(arr)/sizeof(arr[0]); Node* root = insertLevelOrder(arr, 0, n); inOrder(root); } // This code is contributed by Chhavi and improved by Thangaraj |
Java
// Java program to construct binary tree from // given array in level order fashion public class Tree { Node root; // Tree Node static class Node { int data; Node left, right; Node(int data) { this.data = data; this.left = null; this.right = null; } } // Function to insert nodes in level order public Node insertLevelOrder(int[] arr, int i) { Node root = null; // Base case for recursion if (i < arr.length) { root = new Node(arr[i]); // insert left child root.left = insertLevelOrder(arr, 2 * i + 1); // insert right child root.right = insertLevelOrder(arr, 2 * i + 2); } return root; } // Function to print tree nodes in InOrder fashion public void inOrder(Node root) { if (root != null) { inOrder(root.left); System.out.print(root.data + " "); inOrder(root.right); } } // Driver program to test above function public static void main(String args[]) { Tree t2 = new Tree(); int arr[] = { 1, 2, 3, 4, 5, 6, 6, 6, 6 }; t2.root = t2.insertLevelOrder(arr, 0); t2.inOrder(t2.root); } } |
Python3
# Python3 program to construct binary # tree from given array in level # order fashion Tree Node # Helper function that allocates a #new node class newNode: def __init__(self, data): self.data = data self.left = self.right = None # Function to insert nodes in level order def insertLevelOrder(arr, i, n): root = None # Base case for recursion if i < n: root = newNode(arr[i]) # insert left child root.left = insertLevelOrder(arr, 2 * i + 1, n) # insert right child root.right = insertLevelOrder(arr, 2 * i + 2, n) return root # Function to print tree nodes in # InOrder fashion def inOrder(root): if root != None: inOrder(root.left) print(root.data,end=" ") inOrder(root.right) # Driver Code if __name__ == '__main__': arr = [1, 2, 3, 4, 5, 6, 6, 6, 6] n = len(arr) root = None root = insertLevelOrder(arr, 0, n) inOrder(root) # This code is contributed by PranchalK and Improved by Thangaraj |
C#
// C# program to construct binary tree from // given array in level order fashion using System; public class Tree { Node root; // Tree Node public class Node { public int data; public Node left, right; public Node(int data) { this.data = data; this.left = null; this.right = null; } } // Function to insert nodes in level order public Node insertLevelOrder(int[] arr, int i) { Node root = null; // Base case for recursion if (i < arr.Length) { root = new Node(arr[i]); // insert left child root.left = insertLevelOrder(arr, 2 * i + 1); // insert right child root.right = insertLevelOrder(arr, 2 * i + 2); } return root; } // Function to print tree // nodes in InOrder fashion public void inOrder(Node root) { if (root != null) { inOrder(root.left); Console.Write(root.data + " "); inOrder(root.right); } } // Driver code public static void Main(String []args) { Tree t2 = new Tree(); int []arr = { 1, 2, 3, 4, 5, 6, 6, 6, 6 }; t2.root = t2.insertLevelOrder(arr, 0); t2.inOrder(t2.root); } } // This code is contributed Rajput-Ji and improved by Thangaraj |
Javascript
<script> // Javascript program to construct binary tree from // given array in level order fashion let root; class Node { constructor(data) { this.left = null; this.right = null; this.data = data; } } // Function to insert nodes in level order function insertLevelOrder(arr, i) { let root = null; // Base case for recursion if (i < arr.length) { root = new Node(arr[i]); // insert left child root.left = insertLevelOrder(arr, 2 * i + 1); // insert right child root.right = insertLevelOrder(arr, 2 * i + 2); } return root; } // Function to print tree nodes in InOrder fashion function inOrder(root) { if (root != null) { inOrder(root.left); document.write(root.data + " "); inOrder(root.right); } } let arr = [ 1, 2, 3, 4, 5, 6, 6, 6, 6 ]; root = insertLevelOrder(arr, 0); inOrder(root); // This code is contributed by suresh07 and improved by Thangaraj </script> |
Output
6 4 6 2 5 1 6 3 6
Time Complexity: O(n), where n is the total number of nodes in the tree.
Space Complexity: O(n) for calling recursion using stack.
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