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# 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 using namespace std;   // Structure of a node of an n-ary tree struct Node {     int key;     vector 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



Output:

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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