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# Insertion in a Binary Tree in level order

Given a binary tree and a key, insert the key into the binary tree at the first position available in level order.

The idea is to do an iterative level order traversal of the given tree using queue. If we find a node whose left child is empty, we make a new key as the left child of the node. Else if we find a node whose right child is empty, we make the new key as the right child. We keep traversing the tree until we find a node whose either left or right child is empty.

## C++

 // C++ program to insert element in Binary Tree#include #include using namespace std; /* A binary tree node has data, pointer to left childand a pointer to right child */ struct Node {    int data;    Node* left;    Node* right;}; // Function to create a new nodeNode* CreateNode(int data){    Node* newNode = new Node();    if (!newNode) {        cout << "Memory error\n";        return NULL;    }    newNode->data = data;    newNode->left = newNode->right = NULL;    return newNode;} // Function to insert element in binary treeNode* InsertNode(Node* root, int data){    // If the tree is empty, assign new node address to root    if (root == NULL) {        root = CreateNode(data);        return root;    }     // Else, do level order traversal until we find an empty    // place, i.e. either left child or right child of some    // node is pointing to NULL.    queue q;    q.push(root);     while (!q.empty()) {        Node* temp = q.front();        q.pop();         if (temp->left != NULL)            q.push(temp->left);        else {            temp->left = CreateNode(data);            return root;        }         if (temp->right != NULL)            q.push(temp->right);        else {            temp->right = CreateNode(data);            return root;        }    }} /* Inorder traversal of a binary tree */ void inorder(Node* temp){    if (temp == NULL)        return;     inorder(temp->left);    cout << temp->data << ' ';    inorder(temp->right);} // Driver codeint main(){    Node* root = CreateNode(10);    root->left = CreateNode(11);    root->left->left = CreateNode(7);    root->right = CreateNode(9);    root->right->left = CreateNode(15);    root->right->right = CreateNode(8);     cout << "Inorder traversal before insertion: ";    inorder(root);    cout << endl;     int key = 12;    root = InsertNode(root, key);     cout << "Inorder traversal after insertion: ";    inorder(root);    cout << endl;     return 0;}

## Java

 // Java program to insert element in binary treeimport java.util.LinkedList;import java.util.Queue;public class GFG {     /* A binary tree node has key, pointer to    left child and a pointer to right child */    static class Node {        int key;        Node left, right;         // constructor        Node(int key)        {            this.key = key;            left = null;            right = null;        }    }    static Node root;    static Node temp = root;     /* Inorder traversal of a binary tree*/    static void inorder(Node temp)    {        if (temp == null)            return;         inorder(temp.left);        System.out.print(temp.key + " ");        inorder(temp.right);    }     /*function to insert element in binary tree */    static void insert(Node temp, int key)    {         if (temp == null) {            root = new Node(key);            return;        }        Queue q = new LinkedList();        q.add(temp);         // Do level order traversal until we find        // an empty place.        while (!q.isEmpty()) {            temp = q.peek();            q.remove();             if (temp.left == null) {                temp.left = new Node(key);                break;            }            else                q.add(temp.left);             if (temp.right == null) {                temp.right = new Node(key);                break;            }            else                q.add(temp.right);        }    }     // Driver code    public static void main(String args[])    {        root = new Node(10);        root.left = new Node(11);        root.left.left = new Node(7);        root.right = new Node(9);        root.right.left = new Node(15);        root.right.right = new Node(8);         System.out.print(            "Inorder traversal before insertion:");        inorder(root);         int key = 12;        insert(root, key);         System.out.print(            "\nInorder traversal after insertion:");        inorder(root);    }}// This code is contributed by Sumit Ghosh

## Python3

 # Python program to insert element in binary treeclass newNode():     def __init__(self, data):        self.key = data        self.left = None        self.right = None         """ Inorder traversal of a binary tree"""def inorder(temp):     if (not temp):        return     inorder(temp.left)    print(temp.key,end = " ")    inorder(temp.right)  """function to insert element in binary tree """def insert(temp,key):     if not temp:        root = newNode(key)        return    q = []    q.append(temp)     # Do level order traversal until we find    # an empty place.    while (len(q)):        temp = q[0]        q.pop(0)         if (not temp.left):            temp.left = newNode(key)            break        else:            q.append(temp.left)         if (not temp.right):            temp.right = newNode(key)            break        else:            q.append(temp.right)     # Driver codeif __name__ == '__main__':    root = newNode(10)    root.left = newNode(11)    root.left.left = newNode(7)    root.right = newNode(9)    root.right.left = newNode(15)    root.right.right = newNode(8)     print("Inorder traversal before insertion:", end = " ")    inorder(root)     key = 12    insert(root, key)     print()    print("Inorder traversal after insertion:", end = " ")    inorder(root) # This code is contributed by SHUBHAMSINGH10

## C#

 // C# program to insert element in binary treeusing System;using System.Collections.Generic; class GFG {     /* A binary tree node has key, pointer to    left child and a pointer to right child */    public class Node {        public int key;        public Node left, right;         // constructor        public Node(int key)        {            this.key = key;            left = null;            right = null;        }    }    static Node root;     /* Inorder traversal of a binary tree*/    static void inorder(Node temp)    {        if (temp == null)            return;         inorder(temp.left);        Console.Write(temp.key + " ");        inorder(temp.right);    }     /*function to insert element in binary tree */    static void insert(Node temp, int key)    {          if (temp == null) {            root = new Node(key);            return;        }        Queue q = new Queue();        q.Enqueue(temp);         // Do level order traversal until we find        // an empty place.        while (q.Count != 0) {            temp = q.Peek();            q.Dequeue();             if (temp.left == null) {                temp.left = new Node(key);                break;            }            else                q.Enqueue(temp.left);             if (temp.right == null) {                temp.right = new Node(key);                break;            }            else                q.Enqueue(temp.right);        }    }     // Driver code    public static void Main(String[] args)    {        root = new Node(10);        root.left = new Node(11);        root.left.left = new Node(7);        root.right = new Node(9);        root.right.left = new Node(15);        root.right.right = new Node(8);         Console.Write(            "Inorder traversal before insertion:");        inorder(root);         int key = 12;        insert(root, key);         Console.Write(            "\nInorder traversal after insertion:");        inorder(root);    }} // This code is contributed by Rajput-Ji

## Javascript



Output

Inorder traversal before insertion: 7 11 10 15 9 8
Inorder traversal after insertion: 7 11 12 10 15 9 8

Time Complexity:  O(V) where V is the number of nodes.
Auxiliary Space: O(B), where B is the width of the tree and in the worst case we need to hold all vertices of a level in the queue.

This article is contributed by Yash Singla. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.