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: Level 1 (-10): Sum = -10 Level 2 (2, 3): Sum = 5 Level 3 (4, 15, -6): Sum = 13 Level 4 (7, -8, 9): Sum = 8 Only level 3 (4, 15, -6) has sum = K Input: 1 / \ 12 13 / / \ 11 6 -11 \ / 2 2 K = 30 Output: Not Possible Explanation: There is no such level whose sum = K
Approach:
- Perform level order traversal of the Binary tree and store find the sum of each level.
- If the sum is equal to K, print the level. Else move to the next level.
- The process is repeated till all the levels has been traversed and checked.
- If there is no such level with sum K, print “Not Possible”.
Below is the implementation of the above approach:
C++
// C++ program to print all // K-sum levels in a Binary Tree #include <bits/stdc++.h> using namespace std; // Vector to store the // elements of a level vector< int > level; // Binary Tree Node struct node { struct node* left; int data; struct node* right; }; // Function to display elements void display( bool flag) { // Check if boolean variable is true // then print the level if (flag) { for ( auto x : level) cout << x << " " ; } else cout << "Not Possible\n" ; } // Function to find sum of // elements by level order traversal void SumlevelOrder(node* root, int k) { if (root == NULL) return ; // Queue data structure for // level order traversal queue<node*> q; // Enqueue Root in Queue q.push(root); bool flag = false ; while (q.empty() == false ) { // number of nodes at current level int nodeCount = q.size(); int Present_level_sum = 0; // Dequeue all nodes of current level and // Enqueue all nodes of next level while (nodeCount > 0) { node* node = q.front(); // To add node data Present_level_sum += node->data; level.push_back(node->data); q.pop(); if (node->left != NULL) q.push(node->left); if (node->right != NULL) q.push(node->right); nodeCount--; } if (Present_level_sum == k) { flag = true ; break ; } level.clear(); } display(flag); } // Function to create a new tree node node* newNode( int data) { node* temp = new node; temp->data = data; temp->left = NULL; temp->right = NULL; return temp; } // Driver code int main() { // Create binary tree node* root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); root->right->right = newNode(6); int K = 15; SumlevelOrder(root, K); return 0; } |
Java
// Java program to print all // K-sum levels in a Binary Tree import java.util.*; class GFG{ // Vector to store the // elements of a level static Vector<Integer> level = new Vector<Integer>(); // Binary Tree Node static class node { node left; int data; node right; }; // Function to display elements static void display( boolean flag) { // Check if boolean variable is true // then print the level if (flag) { for (Integer x : level) System.out.print(x+ " " ); } else System.out.print( "Not Possible\n" ); } // Function to find sum of // elements by level order traversal static void SumlevelOrder(node root, int k) { if (root == null ) return ; // Queue data structure for // level order traversal Queue<node> q = new LinkedList<>(); // Enqueue Root in Queue q.add(root); boolean flag = false ; while (q.isEmpty() == false ) { // number of nodes at current level int nodeCount = q.size(); int Present_level_sum = 0 ; // Dequeue all nodes of current level and // Enqueue all nodes of next level while (nodeCount > 0 ) { node node = q.peek(); // To add node data Present_level_sum += node.data; level.add(node.data); q.remove(); if (node.left != null ) q.add(node.left); if (node.right != null ) q.add(node.right); nodeCount--; } if (Present_level_sum == k) { flag = true ; break ; } level.clear(); } display(flag); } // Function to create a new tree node static node newNode( int data) { node temp = new node(); temp.data = data; temp.left = null ; temp.right = null ; return temp; } // Driver code public static void main(String[] args) { // Create binary tree node root = newNode( 1 ); root.left = newNode( 2 ); root.right = newNode( 3 ); root.left.left = newNode( 4 ); root.left.right = newNode( 5 ); root.right.right = newNode( 6 ); int K = 15 ; SumlevelOrder(root, K); } } // This code is contributed by 29AjayKumar |
Python3
# Python3 program to print all # K-sum levels in a Binary Tree from collections import deque as queue # A BST node class Node: def __init__( self , x): self .data = x self .left = None self .right = None # Vector to store the # elements of a level level = [] # Function to display elements def display(flag): # Check if boolean variable is # true then print level if (flag): for x in level: print (x, end = " " ) else : print ( "Not Possible\n" ) # Function to find sum of elements # by level order traversal def SumlevelOrder(root, k): if (root = = None ): return # Queue data structure for # level order traversal q = queue() # Enqueue Root in Queue q.append(root) flag = False while ( len (q) > 0 ): # Number of nodes at current level nodeCount = len (q) Present_level_sum = 0 # Dequeue all nodes of current level # and Enqueue all nodes of next level while (nodeCount > 0 ): node = q.popleft() # To add node data Present_level_sum + = node.data level.append(node.data) # q.pop() if (node.left ! = None ): q.append(node.left) if (node.right ! = None ): q.append(node.right) nodeCount - = 1 if (Present_level_sum = = k): flag = True break level.clear() display(flag) # Driver code if __name__ = = '__main__' : # Create binary tree root = Node( 1 ) root.left = Node( 2 ) root.right = Node( 3 ) root.left.left = Node( 4 ) root.left.right = Node( 5 ) root.right.right = Node( 6 ) K = 15 SumlevelOrder(root, K) # This code is contributed by mohit kumar 29 |
C#
// C# program to print all // K-sum levels in a Binary Tree using System; using System.Collections.Generic; class GFG{ // List to store the // elements of a level static List< int > level = new List< int >(); // Binary Tree Node class node { public node left; public int data; public node right; }; // Function to display elements static void display( bool flag) { // Check if bool variable is true // then print the level if (flag) { foreach ( int x in level) Console.Write(x+ " " ); } else Console.Write( "Not Possible\n" ); } // Function to find sum of // elements by level order traversal static void SumlevelOrder(node root, int k) { if (root == null ) return ; // Queue data structure for // level order traversal Queue<node> q = new Queue<node>(); // Enqueue Root in Queue q.Enqueue(root); bool flag = false ; while (q.Count!=0) { // number of nodes at current level int nodeCount = q.Count; int Present_level_sum = 0; // Dequeue all nodes of current level and // Enqueue all nodes of next level while (nodeCount > 0) { node node = q.Peek(); // To add node data Present_level_sum += node.data; level.Add(node.data); q.Dequeue(); if (node.left != null ) q.Enqueue(node.left); if (node.right != null ) q.Enqueue(node.right); nodeCount--; } if (Present_level_sum == k) { flag = true ; break ; } level.Clear(); } display(flag); } // Function to create a new tree node static node newNode( int data) { node temp = new node(); temp.data = data; temp.left = null ; temp.right = null ; return temp; } // Driver code public static void Main(String[] args) { // Create binary tree node root = newNode(1); root.left = newNode(2); root.right = newNode(3); root.left.left = newNode(4); root.left.right = newNode(5); root.right.right = newNode(6); int K = 15; SumlevelOrder(root, K); } } // This code is contributed by sapnasingh4991 |
Javascript
<script> // Javascript program to print all // K-sum levels in a Binary Tree // Vector to store the // elements of a level let level = []; // Binary Tree Node class Node { // Utility function to create a new node constructor(key) { this .data = key; this .left = this .right = null ; } } // Function to display elements function display(flag) { // Check if boolean variable is true // then print the level if (flag) { for (let x = 0; x < level.length; x++) document.write(level[x] + " " ); } else document.write( "Not Possible<br>" ); } // Function to find sum of // elements by level order traversal function SumlevelOrder(root, k) { if (root == null ) return ; // Queue data structure for // level order traversal let q = []; // Enqueue Root in Queue q.push(root); let flag = false ; while (q.length != 0) { // Number of nodes at current level let nodeCount = q.length; let Present_level_sum = 0; // Dequeue all nodes of current level and // Enqueue all nodes of next level while (nodeCount > 0) { let node = q[0]; // To add node data Present_level_sum += node.data; level.push(node.data); q.shift(); if (node.left != null ) q.push(node.left); if (node.right != null ) q.push(node.right); nodeCount--; } if (Present_level_sum == k) { flag = true ; break ; } level = []; } display(flag); } // Driver code let root = new Node(1); root.left = new Node(2); root.right = new Node(3); root.left.left = new Node(4); root.left.right = new Node(5); root.right.right = new Node(6); let K = 15; SumlevelOrder(root, K); // This code is contributed by unknown2108 </script> |
Output:
4 5 6
Time Complexity: O(N) where N is the number of nodes in given binary tree.
Auxiliary Space: O(N) due to queue data structure.
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