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.
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Recursive approach to this problem is discussed here
https://www.geeksforgeeks.org/get-level-of-a-node-in-a-binary-tree/
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
C++
// CPP program to print level of given node// in binary tree iterative approach/* Example binary treeroot is at level 1 20 / \ 10 30 / \ / \ 5 15 25 40 / 12 */#include <bits/stdc++.h>using namespace std;// node of binary treestruct node { int data; node* left; node* right;};// utility function to create// a new nodenode* getnode(int data){ node* newnode = new node(); newnode->data = data; newnode->left = NULL; newnode->right = NULL;}// utility function to return level of given nodeint 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 treeroot 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 poers. 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 nodedef 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 Codeif __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 treeroot 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
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