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.
// 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; } |
Second largest element is : 6
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