Find the Level of a Binary Tree with Width K
Last Updated : 22 Jun, 2021
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 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 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# 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 <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> Time Complexity : O(N)
Auxiliary Space : O(N)
Similar Reads
Maximum width of a Binary Tree with null value
Given a Binary Tree consisting of N nodes, the task is to find the maximum width of the given tree where the maximum width is defined as the maximum of all the widths at each level of the given Tree. The width of a tree for any level is defined as the number of nodes between the two extreme nodes of
11 min read
Maximum width of a Binary Tree with null values | Set 2
Pre-requisite: Maximum width of a Binary Tree with null value | Set 1 Given a Binary Tree consisting of N nodes, the task is to find the maximum width of the given tree without using recursion, where the maximum width is defined as the maximum of all the widths at each level of the given Tree. Note:
8 min read
Level of a Node in Binary Tree
Given a Binary Tree and a key, the task is to find the level of key in the Binary Tree. Examples: Input : key = 4 Output: 3Explanation: The level of the key in above binary tree is 3.Input : key = 10 Output: -1Explanation: Key is not present in the above Binary tree. Table of Content [Expected Appro
12 min read
Nodes at Kth level without duplicates in a Binary Tree
Given a binary tree with N nodes and an integer K, the task is to print nodes of the Kth level of a binary tree without duplicates. Examples: Input: 60 --- Level 0 / \ 50 30 --- Level 1 / \ / 80 10 40 --- Level 2 K = 1 Output: 30 50 Input: 50 --- Level 0 / \ 60 70 --- Level 1 / \ / \ 90 40 40 20 ---
11 min read
Find the closest leaf in a Binary Tree
Given a Binary Tree and a key 'k', find distance of the closest leaf from 'k'. Examples: A / \ B C / \ E F / \ G H / \ / I J K Closest leaf to 'H' is 'K', so distance is 1 for 'H' Closest leaf to 'C' is 'B', so distance is 2 for 'C' Closest leaf to 'E' is either 'I' or 'J', so distance is 2 for 'E'
14 min read
Maximum width of a Binary Tree
Given a binary tree, the task is to find the maximum width of the given tree. The width of a tree is the maximum of the widths of all levels. Before solving the problem first, let us understand what we have to do. Binary trees are one of the most common types of trees in computer science. They are a
15+ min read
Count nodes with two children at level L in a Binary Tree
Given a Binary tree, the task is to count the number of nodes with two children at a given level L. Examples: Input: 1 / \ 2 3 / \ \ 4 5 6 / / \ 7 8 9 L = 2 Output: 1 Input: 20 / \ 8 22 / \ / \ 5 3 4 25 / \ / \ \ 1 10 2 14 6 L = 3 Output: 2 Approach: Initialize a variable count = 0. Recursively trav
13 min read
Print all Nodes of given Binary Tree at the Kth Level
Given a binary tree and an integer K, the task is to print all the integers at the Kth level in the tree from left to right. Examples: Input: Tree in the image below, K = 3 Output: 4 5 6Explanation: All the nodes present in level 3 of above binary tree from left to right are 4, 5, and 6. Input: Tree
5 min read
Check if a binary tree is sorted level-wise or not
Given a binary tree. The task is to check if the binary tree is sorted level-wise or not. A binary tree is level sorted if max( i-1th level) is less than min( ith level ). Examples: Input : 1 / \ / \ 2 3 / \ / \ / \ / \ 4 5 6 7 Output : Sorted Input: 1 / 4 / \ 6 5 \ 2 Output: Not sorted Simple Solut
10 min read
Print all K-sum levels in a Binary Tree
Given a Binary Tree and an integer K where the tree has positive and negative nodes, the task is to print the elements of the level whose sum equals K. If no such result exists, then print "Not Possible". Examples: Input: -10 / \ 2 -3 / \ \ 4 15 -6 / \ / 7 -8 9 K = 13 Output: 4 15 -6 Explanation: Le
8 min read