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Next Larger element in n-ary tree

  • Difficulty Level : Medium
  • Last Updated : 08 Jul, 2021

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;
}

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

Python3




# Python program to find next larger element
# in an n-ary tree.
class Node:
  
# Structure of a node of an n-ary tree
    def __init__(self):
        self.key = 0
        self.child = []
 
# Utility function to create a new tree node
def newNode(key):
    temp = Node()
    temp.key = key
    temp.child = []
    return temp
 
res = None;   
def nextLargerElementUtil(root,x):
    global res
    if (root == None):
        return;
     
     # if root is less than res but
    # greater than x, update res
    if (root.key > x):
        if ((res == None or (res).key > root.key)):
            res = root;
             
    # Number of children of root
    numChildren = len(root.child)
     
     # Recur calling for every child
    for i in range(numChildren):
        nextLargerElementUtil(root.child[i], x)
    return
 
  # Function to find next Greater element
# of x in tree
def nextLargerElement(root,x):
   
   # resultant node
    global res
    res=None
     
    # Calling helper function
    nextLargerElementUtil(root, x)
     
    return res
     
    # Driver code
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))
 
x = 5
print("Next larger element of " , x , " is ",end='')
print(nextLargerElement(root, x).key)
 
# This code is contributed by rag2127.

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

Javascript




<script>
 
// JavaScript program to find next larger element
// in an n-ary tree.
 
// Structure of a node of an n-ary tree
class Node
{
    constructor()
    {
        this.key = 0;
        this.child = [];
    }
};
 
var res = null;
 
// Utility function to create a new tree node
function newNode(key)
{
    var temp = new Node();
    temp.key = key;
    temp.child = [];
    return temp;
}
 
function nextLargerElementUtil(root, 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
    var numChildren = root.child.length;
 
    // Recur calling for every child
    for (var i = 0; i < numChildren; i++)
        nextLargerElementUtil(root.child[i], x);
 
    return;
}
 
// Function to find next Greater element
// of x in tree
function nextLargerElement(root,  x)
{
    // resultant node
    res = null;
 
    // calling helper function
    nextLargerElementUtil(root, x);
 
    return res;
}
 
// Driver Code
/* Let us create below tree
*             5
*         / | \
*         1 2 3
*     / / \ \
*     15 4 5 6
*/
var 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));
var x = 5;
document.write("Next larger element of " +
                             x + " is ");
document.write(nextLargerElement(root, x).key + "<br>");
 
 
</script>

Output:  

Next larger element of 5 is 6

 

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