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# Second Largest element in n-ary tree

• Difficulty Level : Medium
• Last Updated : 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++

 `// 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->child).push_back(newNode(15));``    ``(root->child->child).push_back(newNode(4));``    ``(root->child->child).push_back(newNode(5));``    ``(root->child->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`
Output:
`Second largest element is : 6`

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