Construct a complete binary tree from given array in level order fashion
Last Updated :
09 Jan, 2025
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 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 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# 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
<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>
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;
}
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);
}
}
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
def 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 root
def 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)
using System;
using System.Collections.Generic;
// creating a tree node
public 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);
}
}
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);
Time Complexity: O(n), where n is the total number of nodes in the tree.
Auxiliary Space: O(n)
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