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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://tutorialspoint.dev/slugresolver/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 tree
root is at level 1
  
                20
              /  
            10    30
           /     /
          5  15  25  40
             /
            12  */
#include <iostream>
#include <queue>
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;
}

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 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)

Output:

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

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



This article is attributed to GeeksforGeeks.org

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