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Insertion in a Binary Tree in level order

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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 <iostream>
#include <queue>
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;
    Node* right;
};
 
// Function to create a new node
Node* 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 tree
Node* 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<Node*> 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 code
int 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 tree
import 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<Node> q = new LinkedList<Node>();
        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 tree
class 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 code
if __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 tree
using 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<Node> q = new Queue<Node>();
        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




<script>
// javascript program to insert element in binary tree
 
    /*
     * A binary tree node has key, pointer to left child and a pointer to right
     * child
     */
class Node {
    constructor(val) {
        this.key = val;
        this.left = null;
        this.right = null;
    }
}
 
    var temp;
 
    /* Inorder traversal of a binary tree */
    function inorder(temp) {
        if (temp == null)
            return;
 
        inorder(temp.left);
        document.write(temp.key + " ");
        inorder(temp.right);
    }
 
    /* function to insert element in binary tree */
    function insert(temp , key) {
 
        if (temp == null) {
            root = new Node(key);
            return;
        }
        var q = [];
        q.push(temp);
 
        // Do level order traversal until we find
        // an empty place.
        while (q.length!=0) {
            temp = q.shift();
             
 
            if (temp.left == null) {
                temp.left = new Node(key);
                break;
            } else
                q.push(temp.left);
 
            if (temp.right == null) {
                temp.right = new Node(key);
                break;
            } else
                q.push(temp.right);
        }
    }
 
    // Driver code
        var 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);
 
        document.write("Inorder traversal before insertion:");
        inorder(root);
 
        var key = 12;
        insert(root, key);
 
        document.write("<br\>Inorder traversal after insertion:");
        inorder(root);
 
// This code is contributed by umadevi9616
</script>


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.



Last Updated : 16 Sep, 2022
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