Find the Level of a Binary Tree with Width K

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++

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// 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;
}

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Output:

3

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

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