Second Largest element in n-ary tree
Given an N-ary tree, find and return the node with second largest value in the given tree. Return NULL if no node with required value is present.
For example, in the given tree
Second largest node is 20.
A simple solution is to traverse the array twice. In the first traversal find the maximum value node. In the second traversal find the greatest element node less than the element obtained in first traversal. The time complexity of this solution is O(n).
An Efficient Solution can be to find the second largest element in a single traversal.
Below is the complete algorithm for doing this:
1) Initialize two nodes first and second to NULL as, first = second = NULL 2) Start traversing the tree, a) If the current node data say root->key is greater than first->key then update first and second as, second = first first = root b) If the current node data is in between first and second, then update second to store the value of current node as second = root 3) Return the node stored in second.
Implementation:
C++
// CPP program to find second largest node // in an n-ary tree. #include <bits/stdc++.h> using namespace std; // Structure of a node of an n-ary tree struct Node { int key; vector<Node*> child; }; // Utility function to create a new tree node Node* newNode( int key) { Node* temp = new Node; temp->key = key; return temp; } void secondLargestUtil(Node* root, Node** first, Node** second) { if (root == NULL) return ; // If first is NULL, make root equal to first if (!(*first)) *first = root; // if root is greater than first then second // will become first and update first equal // to root else if (root->key > (*first)->key) { *second = *first; *first = root; } // if second is null, then // update first only if root is less than first else if (!(*second)) { if (root->key < (*first)->key) { *second = root; } } // If root is less than first but greater than second else if (root->key < (*first)->key && root->key > (*second)->key) *second = root; // number of children of root int numChildren = root->child.size(); // recursively calling for every child for ( int i = 0; i < numChildren; i++) secondLargestUtil(root->child[i], first, second); } Node* secondLargest(Node* root) { // second will store the second highest value Node* second = NULL; // first will store the largest value in the tree Node* first = NULL; // calling the helper function secondLargestUtil(root, &first, &second); if (second == NULL) return NULL; // return the second largest element return second; } // Driver program int main() { /* Let us create below tree * 5 * / | \ * 1 2 3 * / / \ \ * 15 4 5 6 */ Node* root = newNode(5); (root->child).push_back(newNode(1)); (root->child).push_back(newNode(2)); (root->child).push_back(newNode(3)); (root->child[0]->child).push_back(newNode(15)); (root->child[1]->child).push_back(newNode(4)); (root->child[1]->child).push_back(newNode(5)); (root->child[2]->child).push_back(newNode(6)); if (secondLargest(root) != NULL) cout << "Second largest element is : " << secondLargest(root)->key << endl; else cout << "Second largest element not found\n" ; return 0; } |
Java
// Java implementation of the approach class GFG { // Class for the node of the tree static class Node { int data; // List of children Node children[]; Node( int n, int data) { children = new Node[n]; this .data = data; } } // Pointers to store the largest and second largest node public static Node largest; public static Node secondLargest; // Helper Function to find the second largest node of the n-ary tree public static void findSecondLargestHelper(Node root) { // Base Case if (root == null ) { return ; } // Check if root's data is larger than current largest node's data if (root.data > largest.data) { secondLargest = largest; largest = root; } else if (root.data > secondLargest.data && root.data != largest.data) secondLargest = root; // recursively find second largest in children for (Node child: root.children) findSecondLargestHelper(child); } // Function to find the second largest node of the n-ary tree public static Node findSecondLargest(Node root) { // Initialising the pointers to a node with value negative infinity largest = new Node( 0 , Integer.MIN_VALUE); secondLargest = largest; findSecondLargestHelper(root); return secondLargest; } // Driver code public static void main(String[] args) { /* Create the following tree 1 / | \ 2 3 4 / | \ 5 6 7 */ int n = 3 ; Node root = new Node(n, 1 ); root.children[ 0 ] = new Node(n, 2 ); root.children[ 1 ] = new Node(n, 3 ); root.children[ 2 ] = new Node(n, 4 ); root.children[ 0 ].children[ 0 ] = new Node(n, 5 ); root.children[ 0 ].children[ 1 ] = new Node(n, 6 ); root.children[ 0 ].children[ 2 ] = new Node(n, 7 ); findSecondLargest(root); System.out.print( "Second Largest Node is: " ); System.out.println(secondLargest.data); } } // This code is contributed by Amitava Mitra |
Python3
# Utility class representing a node of n-ary tree class Node: def __init__( self , key): self .key = key self .child = [] # Utility function to create a new tree node def newNode(key): temp = Node(key) return temp def secondLargestUtil(root, first, second): if root is None : return # If first is None, make root equal to first if first[ 0 ] is None : first[ 0 ] = root # if root is greater than first then second # will become first and update first equal # to root elif root.key > first[ 0 ].key: second[ 0 ] = first[ 0 ] first[ 0 ] = root # if second is None, then # update first only if root is less than first elif second[ 0 ] is None : if root.key < first[ 0 ].key: second[ 0 ] = root # If root is less than first but greater than second elif root.key < first[ 0 ].key and root.key > second[ 0 ].key: second[ 0 ] = root # Recursively calling for every child for i in range ( len (root.child)): secondLargestUtil(root.child[i], first, second) def secondLargest(root): # second will store the second highest value second = [ None ] # first will store the largest value in the tree first = [ None ] # calling the helper function secondLargestUtil(root, first, second) if second[ 0 ] is None : return None # return the second largest element return second[ 0 ] # Driver program if __name__ = = '__main__' : ''' Let us create below tree 5 / | \ 1 2 3 / / \ \ 15 4 5 6 ''' root = newNode( 5 ) root.child.append(newNode( 1 )) root.child.append(newNode( 2 )) root.child.append(newNode( 3 )) root.child[ 0 ].child.append(newNode( 15 )) root.child[ 1 ].child.append(newNode( 4 )) root.child[ 1 ].child.append(newNode( 5 )) root.child[ 2 ].child.append(newNode( 6 )) result = secondLargest(root) if result is not None : print ( "Second largest element is : " , result.key) else : print ( "Second largest element not found" ) # This code is contributed by lokeshpotta20. |
C#
// C# implementation of the approach using System; public class GFG { // Class for the node of the tree public class Node { public int data; // List of children public Node[] children; public Node( int n, int data) { children = new Node[n]; this .data = data; } } // Pointers to store the largest and second largest node static Node largest; static Node secondLargest; // Helper Function to find the second largest node of // the n-ary tree static void findSecondLargestHelper(Node root) { // Base Case if (root == null ) { return ; } // Check if root's data is larger than current // largest node's data if (root.data > largest.data) { secondLargest = largest; largest = root; } else if (root.data > secondLargest.data && root.data != largest.data) secondLargest = root; // recursively find second largest in children foreach (Node child in root.children) findSecondLargestHelper(child); } // Function to find the second largest node of the n-ary // tree static Node findSecondLargest(Node root) { // Initialising the pointers to a node with value // negative infinity largest = new Node(0, Int32.MinValue); secondLargest = largest; findSecondLargestHelper(root); return secondLargest; } // Driver code public static void Main( string [] args) { /* Create the following tree 1 / | \ 2 3 4 / | \ 5 6 7 */ int n = 3; Node root = new Node(n, 1); root.children[0] = new Node(n, 2); root.children[1] = new Node(n, 3); root.children[2] = new Node(n, 4); root.children[0].children[0] = new Node(n, 5); root.children[0].children[1] = new Node(n, 6); root.children[0].children[2] = new Node(n, 7); findSecondLargest(root); Console.Write( "Second Largest Node is: " ); Console.WriteLine(secondLargest.data); } } // This code is contributed by karandeep1234. |
Javascript
<script> class Node { constructor(key) { // key of the node this .key = key; // children of the node this .child = []; } } // Utility function to create a new tree node function newNode(key) { let temp = new Node(key); return temp; } // Helper function for finding the second largest element in the n-ary tree function secondLargestUtil(root, first, second) { // Base case: If root is null, return if (!root) { return ; } // If first is null, make root equal to first if (!first[0]) { first[0] = root; } // If root is greater than first, // then second will become first and update first equal to root else if (root.key > first[0].key) { second[0] = first[0]; first[0] = root; } // If second is null, then update first only if root is less than first else if (!second[0]) { if (root.key < first[0].key) { second[0] = root; } } // If root is less than first but greater than second else if (root.key < first[0].key && root.key > second[0].key) { second[0] = root; } // Recursively calling for every child for (let i = 0; i < root.child.length; i++) { secondLargestUtil(root.child[i], first, second); } } // Function to find the second largest element in the n-ary tree function secondLargest(root) { // second will store the second highest value let second = [ null ]; // first will store the largest value in the tree let first = [ null ]; // calling the helper function secondLargestUtil(root, first, second); // If second is null, return null if (!second[0]) { return null ; } // return the second largest element return second[0]; } // Driver program /* Let us create below tree 5 / | \ 1 2 3 / / \ \ 15 4 5 6 */ let root = newNode(5); root.child.push(newNode(1)); root.child.push(newNode(2)); root.child.push(newNode(3)); root.child[0].child.push(newNode(15)); root.child[1].child.push(newNode(4)); root.child[1].child.push(newNode(5)); root.child[2].child.push(newNode(6)); let result = secondLargest(root); if (result) { document.write( "Second largest element is : " , result.key); } else { document.write( "Second largest element not found" ); } </script> |
Second largest element is : 6
Time Complexity: O(n) where n is the number of nodes in the tree.
Auxiliary Space: O(n)
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