Find the Level of a Binary Tree with Width K

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Given a Binary Tree and an integer K, the task is to find the level of the Binary Tree with width K. If multiple levels exists with width K, print the lowest level. If no such level exists, print -1.

The width of a level of a Binary tree is defined as the number of nodes between leftmost and the rightmost node at that level, including the NULL nodes in between them as well.

Examples:

Input: K = 4

          5  --------- 1st level width = 1 => (5)
        /   \
       6     2 -------- 2nd level width = 2 => (6, 2)
      / \     \  
     7   3     8 -------3rd level width = 4 => (7, 3, NULL, 8)
    /     \
   5       4  -----------4th level width = 4 => (5, NULL, NULL, 4)

Output: 3
Explanation:
For the given tree, the levels having width K( = 4) are 3 and 4.
Since 3 is the minimum of the two, print the minimum.

Input: K = 7

           1  --------- 1st level width = 1 => (1)
         /   \
        2     9 -------- 2nd level width = 2 => (2, 9)
       /       \  
      7         8 ---------3rd level width = 4 => (7, NULL, NULL, 8)
     /         /
    5         9 -----------4th level width = 7 => (5, NULL, NULL, 
             /                                NULL, NULL, NULL, 9)
            2  -----------5th level width = 1 => (2)
           /
          1  -----------6th level width = 1 => (1)

Output: 4
Explanation:
For the given tree, the level having width K( = 7) is 4.

Approach:
The basic idea to solve the problem is to add a label to each node. If a parent has a label i, then assign a label 2*i to it’s left child and 2*i+1 to its right child. This will help in including the NULL nodes in the calculation.
Follow the steps below:

Perform Level Order Traversal on the given tree using a Queue.
Queue contains a pair of {Node, Label}. Initially insert {rootNode, 0} to queue.
If parent has label i, then for a left child, insert {leftChild, 2*i} to queue and for right child, insert{rightChild, 2*i+1} into the queue.
For each level assume a as label of leftmost node and b as label of rightmost node, then (b-a+1) gives the width of that level.
Check whether the width is equal to K. If so, return level.
If none of the levels have width K, then return -1.

Below is the implementation of the above approach:

C++

// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Structure of a Tree node
struct Node {
    int key;
    struct Node *left, *right;
};
 
// Utility function to create
// and initialize a new node
Node* newNode(int key)
{
    Node* temp = new Node;
    temp->key = key;
    temp->left = temp->right = NULL;
    return (temp);
}
 
// Function returns required level
// of width k, if found else -1
int findLevel(Node* root,
              int k, int level)
{
 
    // To store the node and the label
    // and perform traversal
    queue<pair<Node*, int> > qt;
    qt.push(make_pair(root, 0));
 
    int count = 1, b, a = 0;
 
    while (!qt.empty()) {
 
        pair<Node*, int> temp = qt.front();
        qt.pop();
 
        // Taking the last label
        // of each level of the tree
        if (count == 1) {
            b = temp.second;
        }
 
        if ((temp.first)->left) {
            qt.push(make_pair(
                temp.first->left,
                2 * temp.second));
        }
        if (temp.first->right) {
            qt.push(make_pair(
                temp.first->right,
                2 * temp.second + 1));
        }
 
        count--;
 
        // Check width of current level
        if (count == 0) {
 
            // If the width is equal to k
            // then return that level
            if (b - a + 1 == k)
                return level;
 
            pair<Node*, int> secondLabel = qt.front();
 
            // Taking the first label
            // of each level of the tree
            a = secondLabel.second;
 
            level += 1;
            count = qt.size();
        }
    }
 
    // If any level does not has
    // width equal to k, return -1
    return -1;
}
 
// Driver Code
int main()
{
    Node* root = newNode(5);
    root->left = newNode(6);
    root->right = newNode(2);
    root->right->right = newNode(8);
 
    root->left->left = newNode(7);
    root->left->left->left = newNode(5);
    root->left->right = newNode(3);
    root->left->right->right = newNode(4);
 
    int k = 4;
 
    cout << findLevel(root, k, 1) << endl;
 
    return 0;
}

Java

// Java program to implement 
// the above approach 
import java.util.*;
 
class GFG{
 
// Structure of
// binary tree node
static class Node
{
    int data;
    Node left, right;
};
 
static class pair 
{
    Node first;
    int second;
 
    pair(Node first, int second)
    {
        this.first = first;
        this.second = second;
    }
}
 
// Function to create new node
static Node newNode(int data)
{
    Node temp = new Node();
    temp.data = data;
    temp.left = temp.right = null;
    return temp;
}
 
// Function returns required level
// of width k, if found else -1
static int findLevel(Node root,
                     int k, int level)
{
 
    // To store the node and the label
    // and perform traversal
    Queue<pair> qt = new LinkedList<>();
    qt.add(new pair(root, 0));
 
    int count = 1, b = 0, a = 0;
 
    while (!qt.isEmpty()) 
    {
        pair temp = qt.peek();
        qt.poll();
 
        // Taking the last label
        // of each level of the tree
        if (count == 1) 
        {
            b = temp.second;
        }
 
        if (temp.first.left != null) 
        {
            qt.add(new pair(
                temp.first.left,
                2 * temp.second));
        }
        if (temp.first.right != null) 
        {
            qt.add(new pair(
                temp.first.right,
                2 * temp.second + 1));
        }
 
        count--;
 
        // Check width of current level
        if (count == 0) 
        {
             
            // If the width is equal to k
            // then return that level
            if ((b - a + 1) == k)
                return level;
 
            pair secondLabel = qt.peek();
 
            // Taking the first label
            // of each level of the tree
            a = secondLabel.second;
 
            level += 1;
            count = qt.size();
        }
    }
 
    // If any level does not has
    // width equal to k, return -1
    return -1;
}
 
// Driver code
public static void main(String[] args)
{
    Node root = newNode(5);
    root.left = newNode(6);
    root.right = newNode(2);
    root.right.right = newNode(8);
 
    root.left.left = newNode(7);
    root.left.left.left = newNode(5);
    root.left.right = newNode(3);
    root.left.right.right = newNode(4);
 
    int k = 4;
 
    System.out.println(findLevel(root, k, 1));
}
}
 
// This code is contributed by offbeat

Python3

# Python3 program to implement
# the above approach
from collections import deque
 
# Structure of a Tree node
class Node:
     
    def __init__(self, key):
         
        self.key = key
        self.left = None
        self.right = None
 
# Function returns required level
# of width k, if found else -1
def findLevel(root: Node, 
                 k: int, level: int) -> int:
 
    # To store the node and the label
    # and perform traversal
    qt = deque()
    qt.append([root, 0])
 
    count = 1
    b = 0
    a = 0
 
    while qt:
        temp = qt.popleft()
 
        # Taking the last label
        # of each level of the tree
        if (count == 1):
            b = temp[1]
 
        if (temp[0].left):
            qt.append([temp[0].left, 
                   2 * temp[1]])
 
        if (temp[0].right):
            qt.append([temp[0].right, 
                   2 * temp[1] + 1])
 
        count -= 1
 
        # Check width of current level
        if (count == 0):
 
            # If the width is equal to k
            # then return that level
            if (b - a + 1 == k):
                return level
 
            secondLabel = qt[0]
 
            # Taking the first label
            # of each level of the tree
            a = secondLabel[1]
 
            level += 1
            count = len(qt)
 
    # If any level does not has
    # width equal to k, return -1
    return -1
 
# Driver Code
if __name__ == "__main__":
 
    root = Node(5)
    root.left = Node(6)
    root.right = Node(2)
    root.right.right = Node(8)
 
    root.left.left = Node(7)
    root.left.left.left = Node(5)
    root.left.right = Node(3)
    root.left.right.right = Node(4)
 
    k = 4
 
    print(findLevel(root, k, 1))
 
# This code is contributed by sanjeev2552

C#

// C# program to implement 
// the above approach 
using System;
using System.Collections;
using System.Collections.Generic;
  
class GFG{
  
// Structure of
// binary tree node
class Node
{
    public int data;
    public Node left, right;
};
  
class pair 
{
    public Node first;
    public int second;
  
    public pair(Node first, int second)
    {
        this.first = first;
        this.second = second;
    }
}
  
// Function to create new node
static Node newNode(int data)
{
    Node temp = new Node();
    temp.data = data;
    temp.left = temp.right = null;
    return temp;
}
  
// Function returns required level
// of width k, if found else -1
static int findLevel(Node root,
                     int k, int level)
{
  
    // To store the node and the label
    // and perform traversal
    Queue qt = new Queue();
    qt.Enqueue(new pair(root, 0));
  
    int count = 1, b = 0, a = 0;
  
    while (qt.Count!=0) 
    {
        pair temp = (pair)qt.Dequeue();
         
  
        // Taking the last label
        // of each level of the tree
        if (count == 1) 
        {
            b = temp.second;
        }
  
        if (temp.first.left != null) 
        {
            qt.Enqueue(new pair(
                temp.first.left,
                2 * temp.second));
        }
        if (temp.first.right != null) 
        {
            qt.Enqueue(new pair(
                temp.first.right,
                2 * temp.second + 1));
        }
  
        count--;
  
        // Check width of current level
        if (count == 0) 
        {
              
            // If the width is equal to k
            // then return that level
            if ((b - a + 1) == k)
                return level;
  
            pair secondLabel = (pair)qt.Peek();
  
            // Taking the first label
            // of each level of the tree
            a = secondLabel.second;
  
            level += 1;
            count = qt.Count;
        }
    }
  
    // If any level does not has
    // width equal to k, return -1
    return -1;
}
  
// Driver code
public static void Main(string[] args)
{
    Node root = newNode(5);
    root.left = newNode(6);
    root.right = newNode(2);
    root.right.right = newNode(8);
  
    root.left.left = newNode(7);
    root.left.left.left = newNode(5);
    root.left.right = newNode(3);
    root.left.right.right = newNode(4);
  
    int k = 4;
  
    Console.Write(findLevel(root, k, 1));
}
}
 
// This code is contributed by rutvik_56

Javascript

<script>
 
// Javascript program to implement 
// the above approach 
  
// Structure of
// binary tree node
class Node
{
    constructor()
    {
        this.data = 0;
        this.left = null;
        this.right = null;
    }
};
  
class pair 
{
    constructor(first, second)
    {
        this.first = first;
        this.second = second;
    }
}
  
// Function to create new node
function newNode(data)
{
    var temp = new Node();
    temp.data = data;
    temp.left = temp.right = null;
    return temp;
}
  
// Function returns required level
// of width k, if found else -1
function findLevel(root, k, level)
{
  
    // To store the node and the label
    // and perform traversal
    var qt = [];
    qt.push(new pair(root, 0));
  
    var count = 1, b = 0, a = 0;
  
    while (qt.length!=0) 
    {
        var temp = qt.shift();
         
  
        // Taking the last label
        // of each level of the tree
        if (count == 1) 
        {
            b = temp.second;
        }
  
        if (temp.first.left != null) 
        {
            qt.push(new pair(
                temp.first.left,
                2 * temp.second));
        }
        if (temp.first.right != null) 
        {
            qt.push(new pair(
                temp.first.right,
                2 * temp.second + 1));
        }
  
        count--;
  
        // Check width of current level
        if (count == 0) 
        {
              
            // If the width is equal to k
            // then return that level
            if ((b - a + 1) == k)
                return level;
  
            var secondLabel = qt[0];
  
            // Taking the first label
            // of each level of the tree
            a = secondLabel.second;
  
            level += 1;
            count = qt.length;
        }
    }
  
    // If any level does not has
    // width equal to k, return -1
    return -1;
}
  
// Driver code
var root = newNode(5);
root.left = newNode(6);
root.right = newNode(2);
root.right.right = newNode(8);
 
root.left.left = newNode(7);
root.left.left.left = newNode(5);
root.left.right = newNode(3);
root.left.right.right = newNode(4);
 
var k = 4;
 
document.write(findLevel(root, k, 1));
 
 
</script>

Output:

Time Complexity : O(N)
Auxiliary Space : O(N)

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