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 apointer to right child */struct Node{ int data; Node* left, * right;};/* Helper function that allocates anew 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 orderNode* 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 fashionvoid inOrder(Node* root){ if (root != NULL) { inOrder(root->left); cout << root->data <<" "; inOrder(root->right); }}// Driver program to test above functionint 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 fashionpublic 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 nodeclass 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 Codeif __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 fashionusing 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> |
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.
Approach 2: Using queue
Create a TreeNode struct to represent a node in the binary tree.
Define a function buildTree that takes the nums array as a parameter.
If the nums array is empty, return NULL.
Create the root node with the value at index 0 and push it into a queue.
Initialize an integer i to 1.
Loop while the queue is not empty:
Pop the front node from the queue and assign it to curr.
If i is less than the size of the nums array, create a new node with the value at index i and set it as the left child of curr. Increment i by 1. Push the left child node into the queue.
If i is less than the size of the nums array, create a new node with the value at index i and set it as the right child of curr. Increment i by 1. Push the right child node into the queue.
Return the root node.
Define a printTree function to print the values of the tree in preorder traversal order.
Call the buildTree function with the given nums array to construct the complete binary tree.
Call the printTree function to print the values of the tree.
Time complexity: The buildTree function has to visit every element in the nums array once, so the time complexity is O(n), where n is the size of the nums array.
C++
#include <bits/stdc++.h>using namespace std;struct TreeNode { int val; TreeNode *left, *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {}};TreeNode* buildTree(vector<int>& nums) { if (nums.empty()) { return NULL; } TreeNode* root = new TreeNode(nums[0]); queue<TreeNode*> q; q.push(root); int i = 1; while (i < nums.size()) { TreeNode* curr = q.front(); q.pop(); if (i < nums.size()) { curr->left = new TreeNode(nums[i++]); q.push(curr->left); } if (i < nums.size()) { curr->right = new TreeNode(nums[i++]); q.push(curr->right); } } return root;}void printTree(TreeNode* root) { if (!root) { return; } printTree(root->left); cout << root->val << " "; printTree(root->right);}int main() { vector<int> nums = { 1, 2, 3, 4, 5, 6, 6, 6, 6 }; TreeNode* root = buildTree(nums); printTree(root); return 0;} |
Java
import java.util.*;class TreeNode { int val; TreeNode left, right; TreeNode(int x) { val = x; left = null; right = null; }}class Main { public static TreeNode buildTree(int[] nums) { if (nums == null || nums.length == 0) { return null; } TreeNode root = new TreeNode(nums[0]); Queue<TreeNode> q = new LinkedList<>(); q.add(root); int i = 1; while (i < nums.length) { TreeNode curr = q.remove(); if (i < nums.length) { curr.left = new TreeNode(nums[i++]); q.add(curr.left); } if (i < nums.length) { curr.right = new TreeNode(nums[i++]); q.add(curr.right); } } return root; } public static void printTree(TreeNode root) { if (root == null) { return; } printTree(root.left); System.out.print(root.val + " "); printTree(root.right); } public static void main(String[] args) { int[] nums = { 1, 2, 3, 4, 5, 6, 6, 6, 6 }; TreeNode root = buildTree(nums); printTree(root); }} |
Python3
class TreeNode: def __init__(self, x): self.val = x self.left = None self.right = Nonedef buildTree(nums): if not nums: return None root = TreeNode(nums[0]) q = [root] i = 1 while i < len(nums): curr = q.pop(0) if i < len(nums): curr.left = TreeNode(nums[i]) q.append(curr.left) i += 1 if i < len(nums): curr.right = TreeNode(nums[i]) q.append(curr.right) i += 1 return rootdef printTree(root): if not root: return printTree(root.left) print(root.val, end=" ") printTree(root.right)nums = [1, 2, 3, 4, 5, 6, 6, 6, 6]root = buildTree(nums)printTree(root) |
C#
using System;using System.Collections.Generic;// creating a tree nodepublic class TreeNode{ public int val; public TreeNode left, right; public TreeNode(int x) { val = x; left = null; right = null; }}public class MainClass{ public static TreeNode BuildTree(int[] nums) { if (nums == null || nums.Length == 0) { return null; } TreeNode root = new TreeNode(nums[0]); // taking a queue Queue<TreeNode> q = new Queue<TreeNode>(); q.Enqueue(root); int i = 1; while (i < nums.Length) { TreeNode curr = q.Dequeue(); if (i < nums.Length) { // traversing left curr.left = new TreeNode(nums[i++]); q.Enqueue(curr.left); } if (i < nums.Length) { // traversing right curr.right = new TreeNode(nums[i++]); q.Enqueue(curr.right); } } return root; } public static void PrintTree(TreeNode root) { if (root == null) { return; } PrintTree(root.left); Console.Write(root.val + " "); PrintTree(root.right); } // Driver code public static void Main(string[] args) { int[] nums = { 1, 2, 3, 4, 5, 6, 6, 6, 6 }; TreeNode root = BuildTree(nums); PrintTree(root); }} |
Javascript
class TreeNode { constructor(val) { this.val = val; this.left = null; this.right = null; }}function buildTree(nums) { if (nums.length === 0) { return null; } let root = new TreeNode(nums[0]); let q = [root]; let i = 1; while (i < nums.length) { let curr = q.shift(); if (i < nums.length) { curr.left = new TreeNode(nums[i++]); q.push(curr.left); } if (i < nums.length) { curr.right = new TreeNode(nums[i++]); q.push(curr.right); } } return root;}function printTree(root) { if (!root) { return; } printTree(root.left); console.log(root.val + " "); printTree(root.right);}let nums = [1, 2, 3, 4, 5, 6, 6, 6, 6];let root = buildTree(nums);printTree(root); |
6 4 6 2 5 1 6 3 6
Time Complexity: O(n), where n is the total number of nodes in the tree.
Auxiliary Space: O(n)
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