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.
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 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
// 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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