Get Level of a node in a Binary Tree

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.

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

The idea is to start from the root and level as 1. If the key matches with root’s data, return level. Else recursively call for left and right subtrees with level as level + 1.

C++

// C++ program to Get Level of a  
// node in a Binary Tree 
#include<bits/stdc++.h> 
using namespace std; 
  
/* A tree node structure */
struct node 
{ 
    int data; 
    struct node *left; 
    struct node *right; 
}; 
  
/* Helper function for getLevel().  
It returns level of the data if data is 
present in tree, otherwise returns 0.*/
int getLevelUtil(struct node *node,  
                 int data, int level) 
{ 
    if (node == NULL) 
        return 0; 
  
    if (node -> data == data) 
        return level; 
  
    int downlevel = getLevelUtil(node -> left,  
                                 data, level + 1); 
    if (downlevel != 0) 
        return downlevel; 
  
    downlevel = getLevelUtil(node->right,  
                             data, level + 1); 
    return downlevel; 
} 
  
/* Returns level of given data value */
int getLevel(struct node *node, int data) 
{ 
    return getLevelUtil(node, data, 1); 
} 
  
/* Utility function to create a  
new Binary Tree node */
struct node* newNode(int data) 
{ 
    struct node *temp = new struct node; 
    temp -> data = data; 
    temp -> left = NULL; 
    temp -> right = NULL; 
  
    return temp; 
} 
  
// Driver Code 
int main() 
{ 
    struct node *root = new struct node; 
    int x; 
  
    /* Constructing tree given in  
    the above figure */
    root = newNode(3); 
    root->left = newNode(2); 
    root->right = newNode(5); 
    root->left->left = newNode(1); 
    root->left->right = newNode(4); 
  
    for (x = 1; x <= 5; x++) 
    { 
        int level = getLevel(root, x); 
        if (level) 
            cout << "Level of "<< x << " is "
                 << getLevel(root, x) << endl; 
        else
            cout << x << "is not present in tree"
                      << endl; 
    } 
  
    getchar(); 
    return 0; 
} 
  
// This code is contributed 
// by Akanksha Rai 

C

#include<stdio.h> 
// C program to Get Level of a node in a Binary Tree 
/* A tree node structure */
struct node 
{ 
    int data; 
    struct node *left; 
    struct node *right; 
}; 
  
/* Helper function for getLevel().  It returns level of the data if data is 
   present in tree, otherwise returns 0.*/
int getLevelUtil(struct node *node, int data, int level) 
{ 
    if (node == NULL) 
        return 0; 
  
    if (node->data == data) 
        return level; 
  
    int downlevel = getLevelUtil(node->left, data, level+1); 
    if (downlevel != 0) 
        return downlevel; 
  
    downlevel = getLevelUtil(node->right, data, level+1); 
    return downlevel; 
} 
  
/* Returns level of given data value */
int getLevel(struct node *node, int data) 
{ 
    return getLevelUtil(node,data,1); 
} 
  
/* Utility function to create a new Binary Tree node */
struct node* newNode(int data) 
{ 
    struct node *temp = new struct node; 
    temp->data = data; 
    temp->left = NULL; 
    temp->right = NULL; 
  
    return temp; 
} 
  
/* Driver function to test above functions */
int main() 
{ 
    struct node *root = new struct node; 
    int x; 
  
    /* Constructing tree given in the above figure */
    root = newNode(3); 
    root->left = newNode(2); 
    root->right = newNode(5); 
    root->left->left = newNode(1); 
    root->left->right = newNode(4); 
  
    for (x = 1; x <=5; x++) 
    { 
      int level = getLevel(root, x); 
      if (level) 
        printf(" Level of %d is %d\n", x, getLevel(root, x)); 
      else
        printf(" %d is not present in tree \n", x); 
  
    } 
  
    getchar(); 
    return 0; 
} 

Java

// Java program to Get Level of a node in a Binary Tree 
/* A tree node structure */
class Node  
{ 
    int data; 
    Node left, right; 
   
    public Node(int d)  
    { 
        data = d; 
        left = right = null; 
    } 
} 
   
class BinaryTree  
{ 
   
    Node root; 
   
    /* Helper function for getLevel().  It returns level of the data 
    if data is present in tree, otherwise returns 0.*/
    int getLevelUtil(Node node, int data, int level)  
    { 
        if (node == null) 
            return 0; 
   
        if (node.data == data) 
            return level; 
   
        int downlevel = getLevelUtil(node.left, data, level + 1); 
        if (downlevel != 0) 
            return downlevel; 
   
        downlevel = getLevelUtil(node.right, data, level + 1); 
        return downlevel; 
    } 
   
    /* Returns level of given data value */
    int getLevel(Node node, int data)  
    { 
        return getLevelUtil(node, data, 1); 
    } 
   
    /* Driver function to test above functions */
    public static void main(String[] args)  
    { 
        BinaryTree tree = new BinaryTree(); 
   
        /* Constructing tree given in the above figure */
        tree.root = new Node(3); 
        tree.root.left = new Node(2); 
        tree.root.right = new Node(5); 
        tree.root.left.left = new Node(1); 
        tree.root.left.right = new Node(4); 
        for (int x = 1; x <= 5; x++)  
        { 
            int level = tree.getLevel(tree.root, x); 
            if (level != 0) 
                System.out.println("Level of " + x + " is "
                        + tree.getLevel(tree.root, x)); 
            else
                System.out.println(x + " is not present in tree"); 
        } 
    } 
} 
   
// This code has been contributed by Mayank Jaiswal(mayank_24) 

Python3

# Python3 program to Get Level of a 
# node in a Binary Tree 
  
# Helper function that allocates a   
# new node with the given data and  
# None left and right pairs.                                      
class newNode:  
  
    # Constructor to create a new node  
    def __init__(self, data):  
        self.data = data 
        self.left = None
        self.right = None
  
# Helper function for getLevel(). It  
# returns level of the data if data is  
# present in tree, otherwise returns 0 
def getLevelUtil(node, data, level): 
    if (node == None): 
        return 0
  
    if (node.data == data) : 
        return level  
  
    downlevel = getLevelUtil(node.left, 
                             data, level + 1)  
    if (downlevel != 0) : 
        return downlevel  
  
    downlevel = getLevelUtil(node.right,  
                             data, level + 1)  
    return downlevel  
  
# Returns level of given data value  
def getLevel(node, data) : 
  
    return getLevelUtil(node, data, 1)  
  
# Driver Code  
if __name__ == '__main__': 
      
    # Let us construct the Tree shown  
    # in the above figure  
    root = newNode(3)  
    root.left = newNode(2)  
    root.right = newNode(5)  
    root.left.left = newNode(1)  
    root.left.right = newNode(4) 
    for x in range(1, 6): 
        level = getLevel(root, x) 
        if (level) : 
            print("Level of", x,  
                  "is", getLevel(root, x)) 
        else: 
            print(x, "is not present in tree")  
  
# This code is contributed by 
# Shubham Singh(SHUBHAMSINGH10) 

C#

// C# program to Get Level of a node in a Binary Tree 
using System; 
  
/* A tree node structure */
public class Node 
{ 
    public int data; 
    public Node left, right; 
  
    public Node(int d) 
    { 
        data = d; 
        left = right = null; 
    } 
} 
  
public class BinaryTree 
{ 
  
    public Node root; 
  
    /* Helper function for getLevel().  It returns level of the data  
    if data is present in tree, otherwise returns 0.*/
    public virtual int getLevelUtil(Node node, int data, int level) 
    { 
        if (node == null) 
        { 
            return 0; 
        } 
  
        if (node.data == data) 
        { 
            return level; 
        } 
  
        int downlevel = getLevelUtil(node.left, data, level + 1); 
        if (downlevel != 0) 
        { 
            return downlevel; 
        } 
  
        downlevel = getLevelUtil(node.right, data, level + 1); 
        return downlevel; 
    } 
  
    /* Returns level of given data value */
    public virtual int getLevel(Node node, int data) 
    { 
        return getLevelUtil(node, data, 1); 
    } 
  
    /* Driver function to test above functions */
    public static void Main(string[] args) 
    { 
        BinaryTree tree = new BinaryTree(); 
  
        /* Constructing tree given in the above figure */
        tree.root = new Node(3); 
        tree.root.left = new Node(2); 
        tree.root.right = new Node(5); 
        tree.root.left.left = new Node(1); 
        tree.root.left.right = new Node(4); 
        for (int x = 1; x <= 5; x++) 
        { 
            int level = tree.getLevel(tree.root, x); 
            if (level != 0) 
            { 
                Console.WriteLine("Level of " + x + " is " + tree.getLevel(tree.root, x)); 
            } 
            else
            { 
                Console.WriteLine(x + " is not present in tree"); 
            } 
        } 
    } 
} 
  
  // This code is contributed by Shrikant13 

Output:

 Level of 1 is 3
 Level of 2 is 2
 Level of 3 is 1
 Level of 4 is 3
 Level of 5 is 2

Time Complexity of getLevel() is O(n) where n is the number of nodes in the given Binary Tree.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

C

Java

Python3

C#

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