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>

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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