Second Largest element in n-ary tree

09 Jul, 2021
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.
 
C++
Java
// 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 treestruct Node {    int key;    vector<Node*> child;}; // Utility function to create a new tree nodeNode* 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 programint 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 implementation of the approachclass 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
Second largest element is : 6
 
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