Get level of a node in binary tree | iterative approach

Given a Binary Tree and a key, write a function that returns level of the key.

For example, consider the following tree. If the input key is 3, then your function should return 1. If the input key is 4, then your function should return 3. And for key which is not present in key, then your function should return 0.

Recursive approach to this problem is discussed here 
https://www.geeksforgeeks.org/get-level-of-a-node-in-a-binary-tree/amp/

The iterative approach is discussed below : 
The iterative approach is modified version of Level Order Tree Traversal

Algorithm

create a empty queue q
push root and then NULL to q
loop till q is not empty
   get the front node into temp node
   if it is NULL, it means all nodes of 
      one level are traversed, so increment 
      level
   else 
     check if temp data is equal to data  
     to be searched
     if yes then return level
     if temp->left is not NULL, 
         enqueue temp->left
     if temp->right is not NULL, 
         enqueue temp->right
if value not found, then return 0

Implementation:

C++

// CPP program to print level of given node
// in binary tree iterative approach
/* Example binary tree
root is at level 1

                20
              /   \
            10    30
           / \    / \
          5  15  25  40
             /
            12  */
#include <bits/stdc++.h>
using namespace std;

// node of binary tree
struct node {
    int data;
    node* left;
    node* right;
};

// utility function to create
// a new node
node* getnode(int data)
{
    node* newnode = new node();
    newnode->data = data;
    newnode->left = NULL;
    newnode->right = NULL;
}

// utility function to return level of given node
int getlevel(node* root, int data)
{
    queue<node*> q;
    int level = 1;
    q.push(root);

    // extra NULL is pushed to keep track
    // of all the nodes to be pushed before
    // level is incremented by 1
    q.push(NULL);
    while (!q.empty()) {
        node* temp = q.front();
        q.pop();
        if (temp == NULL) {
            if (q.front() != NULL) {
                q.push(NULL);
            }
            level += 1;
        } else {
            if (temp->data == data) {
                return level;
            }
            if (temp->left) {
                q.push(temp->left);
            }
            if (temp->right) {
                q.push(temp->right);
            }
        }
    }
    return 0;
}

int main()
{
    // create a binary tree
    node* root = getnode(20);
    root->left = getnode(10);
    root->right = getnode(30);
    root->left->left = getnode(5);
    root->left->right = getnode(15);
    root->left->right->left = getnode(12);
    root->right->left = getnode(25);
    root->right->right = getnode(40);

    // return level of node
    int level = getlevel(root, 30);
    (level != 0) ? (cout << "level of node 30 is " << level << endl) :
                   (cout << "node 30 not found" << endl);

    level = getlevel(root, 12);
    (level != 0) ? (cout << "level of node 12 is " << level << endl) :
                   (cout << "node 12 not found" << endl);

    level = getlevel(root, 25);
    (level != 0) ? (cout << "level of node 25 is " << level << endl) :
                   (cout << "node 25 not found" << endl);

    level = getlevel(root, 27);
    (level != 0) ? (cout << "level of node 27 is " << level << endl) :
                   (cout << "node 27 not found" << endl);
    return 0;
}

Java

// Java program to print level of given node
// in binary tree iterative approach
/* Example binary tree
root is at level 1

                20
            / \
            10 30
        / \ / \
        5 15 25 40
            /
            12 */
import java.io.*;
import java.util.*;

class GFG
{

    // node of binary tree
    static class node
    {
        int data;
        node left, right;

        node(int data)
        {
            this.data = data;
            this.left = this.right = null;
        }
    }

    // utility function to return level of given node
    static int getLevel(node root, int data)
    {
        Queue<node> q = new LinkedList<>();
        int level = 1;
        q.add(root);

        // extra NULL is pushed to keep track
        // of all the nodes to be pushed before
        // level is incremented by 1
        q.add(null);
        while (!q.isEmpty())
        {
            node temp = q.poll();
            if (temp == null)
            {
                if (q.peek() != null)
                {
                    q.add(null);
                }
                level += 1;
            }
            else
            {
                if (temp.data == data)
                {
                    return level;
                }
                if (temp.left != null)
                {
                    q.add(temp.left);
                }
                if (temp.right != null)
                {
                    q.add(temp.right);
                }
            }
        }
        return 0;
    }

    // Driver Code
    public static void main(String[] args)
    {

        // create a binary tree
        node root = new node(20);
        root.left = new node(10);
        root.right = new node(30);
        root.left.left = new node(5);
        root.left.right = new node(15);
        root.left.right.left = new node(12);
        root.right.left = new node(25);
        root.right.right = new node(40);

        // return level of node
        int level = getLevel(root, 30);
        if (level != 0)
            System.out.println("level of node 30 is " + level);
        else
            System.out.println("node 30 not found");

        level = getLevel(root, 12);
        if (level != 0)
            System.out.println("level of node 12 is " + level);
        else
            System.out.println("node 12 not found");

        level = getLevel(root, 25);
        if (level != 0)
            System.out.println("level of node 25 is " + level);
        else
            System.out.println("node 25 not found");

        level = getLevel(root, 27);
        if (level != 0)
            System.out.println("level of node 27 is " + level);
        else
            System.out.println("node 27 not found");
    }
}

// This code is contributed by
// sanjeev2552

Python3

# Python3 program to find closest
# value in Binary search Tree

_MIN = -2147483648
_MAX = 2147483648

# Helper function that allocates a new
# node with the given data and None
# left and right pointers.
class getnode:

    # Constructor to create a new node
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None

# utility function to return level
# of given node
def getlevel(root, data):

    q = []
    level = 1
    q.append(root)

    # extra None is appended to keep track
    # of all the nodes to be appended
    # before level is incremented by 1
    q.append(None)
    while (len(q)):
        temp = q[0]
        q.pop(0)
        if (temp == None) :
            if len(q) == 0:
                return 0
            if (q[0] != None):
                q.append(None)
            level += 1
        else :
            if (temp.data == data) :
                return level
            if (temp.left):
                q.append(temp.left)
            if (temp.right) :
                q.append(temp.right)
    return 0

# Driver Code
if __name__ == '__main__':

    # create a binary tree
    root = getnode(20)
    root.left = getnode(10)
    root.right = getnode(30)
    root.left.left = getnode(5)
    root.left.right = getnode(15)
    root.left.right.left = getnode(12)
    root.right.left = getnode(25)
    root.right.right = getnode(40)

    # return level of node
    level = getlevel(root, 30)
    if level != 0:
        print("level of node 30 is", level)
    else:
        print("node 30 not found")

    level = getlevel(root, 12)
    if level != 0:
        print("level of node 12 is", level)
    else:
        print("node 12 not found")

    level = getlevel(root, 25)
    if level != 0:
        print("level of node 25 is", level)
    else:
        print("node 25 not found")

    level = getlevel(root, 27)
    if level != 0:
        print("level of node 27 is", level)
    else:
        print("node 27 not found")

# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)

C#

// C# program to print level of given node
// in binary tree iterative approach
/* Example binary tree
root is at level 1

                20
            / \
            10 30
        / \ / \
        5 15 25 40
            /
            12 */
using System;
using System.Collections;
using System.Collections.Generic;

class GFG
{

    // node of binary tree
    public class node
    {
        public int data;
        public node left, right;

        public node(int data)
        {
            this.data = data;
            this.left = this.right = null;
        }
    }

    // utility function to return level of given node
    static int getLevel(node root, int data)
    {
        Queue<node> q = new Queue<node>();
        int level = 1;
        q.Enqueue(root);

        // extra NULL is pushed to keep track
        // of all the nodes to be pushed before
        // level is incremented by 1
        q.Enqueue(null);
        while (q.Count > 0)
        {
            node temp = q.Dequeue();

            if (temp == null)
            {
                if (q.Count > 0)
                {
                    q.Enqueue(null);
                }
                level += 1;
            }
            else
            {
                if (temp.data == data)
                {
                    return level;
                }
                if (temp.left != null)
                {
                    q.Enqueue(temp.left);
                }
                if (temp.right != null)
                {
                    q.Enqueue(temp.right);
                }
            }
        }
        return 0;
    }

    // Driver Code
    public static void Main(String []args)
    {

        // create a binary tree
        node root = new node(20);
        root.left = new node(10);
        root.right = new node(30);
        root.left.left = new node(5);
        root.left.right = new node(15);
        root.left.right.left = new node(12);
        root.right.left = new node(25);
        root.right.right = new node(40);

        // return level of node
        int level = getLevel(root, 30);
        if (level != 0)
            Console.WriteLine("level of node 30 is " + level);
        else
            Console.WriteLine("node 30 not found");

        level = getLevel(root, 12);
        if (level != 0)
            Console.WriteLine("level of node 12 is " + level);
        else
            Console.WriteLine("node 12 not found");

        level = getLevel(root, 25);
        if (level != 0)
            Console.WriteLine("level of node 25 is " + level);
        else
            Console.WriteLine("node 25 not found");

        level = getLevel(root, 27);
        if (level != 0)
            Console.WriteLine("level of node 27 is " + level);
        else
            Console.WriteLine("node 27 not found");
    }
}

// This code is contributed by Arnab Kundu

Javascript

<script>
    // Javascript program to print level of given node
    // in binary tree iterative approach
    /* Example binary tree
    root is at level 1

                    20
                / \
                10 30
            / \ / \
            5 15 25 40
                /
                12 */

    class node
    {
        constructor(data) {
           this.left = null;
           this.right = null;
           this.data = data;
        }
    }

      // utility function to return level of given node
    function getLevel(root, data)
    {
        let q = [];
        let level = 1;
        q.push(root);

        // extra NULL is pushed to keep track
        // of all the nodes to be pushed before
        // level is incremented by 1
        q.push(null);
        while (q.length > 0)
        {
            let temp = q[0];
            q.shift()
            if (temp == null)
            {
                if (q[0] != null)
                {
                    q.push(null);
                }
                level += 1;
            }
            else
            {
                if (temp.data == data)
                {
                    return level;
                }
                if (temp.left != null)
                {
                    q.push(temp.left);
                }
                if (temp.right != null)
                {
                    q.push(temp.right);
                }
            }
        }
        return 0;
    }

    // create a binary tree
    let root = new node(20);
    root.left = new node(10);
    root.right = new node(30);
    root.left.left = new node(5);
    root.left.right = new node(15);
    root.left.right.left = new node(12);
    root.right.left = new node(25);
    root.right.right = new node(40);

    // return level of node
    let level = getLevel(root, 30);
    if (level != 0)
      document.write("level of node 30 is " + level + "</br>");
    else
      document.write("node 30 not found" + "</br>");

    level = getLevel(root, 12);
    if (level != 0)
      document.write("level of node 12 is " + level + "</br>");
    else
      document.write("node 12 not found" + "</br>");

    level = getLevel(root, 25);
    if (level != 0)
      document.write("level of node 25 is " + level + "</br>");
    else
      document.write("node 25 not found" + "</br>");

    level = getLevel(root, 27);
    if (level != 0)
      document.write("level of node 27 is " + level + "</br>");
    else
      document.write("node 27 not found" + "</br>");

    // This code is contributed by suresh07.
</script>

Output

level of node 30 is 2
level of node 12 is 4
level of node 25 is 3
node 27 not found

Time complexity: O(n) // where n is no of nodes in given Binary Tree

Auxiliary space: O(n)

https://www.youtube.com/watch?v=MHiHfW4iPlk