Next Larger element in n-ary tree
Given a generic tree and a integer x. Find and return the node with next larger element in the tree i.e. find a node just greater than x. Return NULL if no node is present with value greater than x.
For example, in the given tree
x = 10, just greater node value is 12
The idea is maintain a node pointer res, which will contain the final resultant node.
Traverse the tree and check if root data is greater than x. If so, then compare the root data with res data.
If root data is greater than n and less than res data update res.
C++
// CPP program to find next larger element // 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 nextLargerElementUtil(Node* root, int x, Node** res) { if (root == NULL) return; // if root is less than res but greater than // x update res if (root->key > x) if (!(*res) || (*res)->key > root->key) *res = root; // Number of children of root int numChildren = root->child.size(); // Recur calling for every child for (int i = 0; i < numChildren; i++) nextLargerElementUtil(root->child[i], x, res); return; } // Function to find next Greater element of x in tree Node* nextLargerElement(Node* root, int x) { // resultant node Node* res = NULL; // calling helper function nextLargerElementUtil(root, x, &res); return res; } // 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)); int x = 5; cout << "Next larger element of " << x << " is "; cout << nextLargerElement(root, x)->key << endl; return 0; } |
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Java
// Java program to find next larger element // in an n-ary tree. import java.util.*; class GFG { // Structure of a node of an n-ary tree static class Node { int key; Vector<Node> child; }; static Node res; // Utility function to create a new tree node static Node newNode(int key) { Node temp = new Node(); temp.key = key; temp.child = new Vector<>(); return temp; } static void nextLargerElementUtil(Node root, int x) { if (root == null) return; // if root is less than res but // greater than x, update res if (root.key > x) if ((res == null || (res).key > root.key)) res = root; // Number of children of root int numChildren = root.child.size(); // Recur calling for every child for (int i = 0; i < numChildren; i++) nextLargerElementUtil(root.child.get(i), x); return; } // Function to find next Greater element // of x in tree static Node nextLargerElement(Node root, int x) { // resultant node res = null; // calling helper function nextLargerElementUtil(root, x); return res; } // Driver Code public static void main(String[] args) { /* Let us create below tree * 5 * / | \ * 1 2 3 * / / \ \ * 15 4 5 6 */ Node root = newNode(5); (root.child).add(newNode(1)); (root.child).add(newNode(2)); (root.child).add(newNode(3)); (root.child.get(0).child).add(newNode(15)); (root.child.get(1).child).add(newNode(4)); (root.child.get(1).child).add(newNode(5)); (root.child.get(2).child).add(newNode(6)); int x = 5; System.out.print("Next larger element of " + x + " is "); System.out.print(nextLargerElement(root, x).key + "\n"); } } // This code is contributed by 29AjayKumar |
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C#
// C# program to find next larger element // in an n-ary tree. using System; using System.Collections.Generic; class GFG { // Structure of a node of an n-ary tree class Node { public int key; public List<Node> child; }; static Node res; // Utility function to create a new tree node static Node newNode(int key) { Node temp = new Node(); temp.key = key; temp.child = new List<Node>(); return temp; } static void nextLargerElementUtil(Node root, int x) { if (root == null) return; // if root is less than res but // greater than x, update res if (root.key > x) if ((res == null || (res).key > root.key)) res = root; // Number of children of root int numChildren = root.child.Count; // Recur calling for every child for (int i = 0; i < numChildren; i++) nextLargerElementUtil(root.child[i], x); return; } // Function to find next Greater element // of x in tree static Node nextLargerElement(Node root, int x) { // resultant node res = null; // calling helper function nextLargerElementUtil(root, x); return res; } // Driver Code public static void Main(String[] args) { /* Let us create below tree * 5 * / | \ * 1 2 3 * / / \ \ * 15 4 5 6 */ Node root = newNode(5); (root.child).Add(newNode(1)); (root.child).Add(newNode(2)); (root.child).Add(newNode(3)); (root.child[0].child).Add(newNode(15)); (root.child[1].child).Add(newNode(4)); (root.child[1].child).Add(newNode(5)); (root.child[2].child).Add(newNode(6)); int x = 5; Console.Write("Next larger element of " + x + " is "); Console.Write(nextLargerElement(root, x).key + "\n"); } } // This code is contributed by PrinciRaj1992 |
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Output:
Next larger element of 5 is 6
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