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.

Implementation:

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

Example no2:

Algorithmic steps:

Traverse the tree in post-order and store the nodes in a stack.
Create an empty hash map to store the next larger element for each node.
For each node in the stack, pop it from the stack and find the next larger element for it by checking the elements on the top of the stack. If an element on the top of the stack is greater than the current node, it is the next larger element for the current node. Keep popping elements from the stack until the top of the stack is less than or equal to the current node.
Add the next larger element for the current node to the hash map.
Repeat steps 3 and 4 until the stack is empty.
Return the next larger element for the target node by looking it up in the hash map.

Note: This algorithm assumes that the tree is a binary search tree, where the value of each node is greater than the values of its left child and less than the values of its right child. If the tree is not a binary search tree, the above algorithm may not give correct result

Program: Implementing by Postorder traversing tree:

C++

#include <iostream>
#include <stack>
#include <unordered_map>

struct TreeNode {
    int val;
    vector<TreeNode*> children;
    TreeNode(int x) : val(x) {}
};

void postOrder(TreeNode *root, stack<int> &nodes) {
    if (!root) return;
    for (int i = 0; i < root->children.size(); i++) {
        postOrder(root->children[i], nodes);
    }
    nodes.push(root->val);
}

unordered_map<int, int> findNextLarger(TreeNode *root) {
    stack<int> nodes;
    unordered_map<int, int> nextLarger;
    postOrder(root, nodes);

    while (!nodes.empty()) {
        int current = nodes.top();
        nodes.pop();
        while (!nodes.empty() && nodes.top() <= current) {
            nodes.pop();
        }
        if (!nodes.empty()) {
            nextLarger[current] = nodes.top();
        } else {
            nextLarger[current] = -1;
        }
    }

    return nextLarger;
}

int main() {
    TreeNode *root = new TreeNode(8);
    root->children.push_back(new TreeNode(3));
    root->children.push_back(new TreeNode(10));
    root->children[0]->children.push_back(new TreeNode(1));
    root->children[0]->children.push_back(new TreeNode(6));
    root->children[0]->children[1]->children.push_back(new TreeNode(4));
    root->children[0]->children[1]->children.push_back(new TreeNode(7));
    root->children[1]->children.push_back(new TreeNode(14));

    int target = 4;
    unordered_map<int, int> nextLarger = findNextLarger(root);
    cout << "The next larger element for " << target << " is: " << nextLarger[target] << endl;
    return 0;
}

The next larger element for 4 is 7

Time and Space complexities are:

The time complexity of this algorithm is O(n), where n is the number of nodes in the n-ary tree. The reason for this is that we traverse each node in the tree once, and for each node, we perform a constant amount of work to find its next larger element.

The auxiliary space of this algorithm is O(n), where n is the number of nodes in the n-ary tree. The reason for this is that we use a stack to store the nodes in post-order and a hash map to store the next larger element for each node. The size of the stack and the hash map is proportional to the number of nodes in the tree, so their combined space complexity is O(n).

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