Kth Smallest Element in an N-ary Tree

Last Updated : 23 Jul, 2025

Given an N-array Tree (Generic Tree) and an integer K, the task is to find the K^th smallest element in an N-array Tree.

Examples:

Input: 10
/ / \ \
2 34 56 100
/ \ | / | \
77 88 1 7 8 9
K = 3
Output: 7
Explanation: 7 is the 3rd smallest element in the tree. The first two smallest elements are 1 and 2 respectively.

Input: 1
/ \ \
2 3 4
/ \
5 6
K = 4
Output: 4

Approach: The problem can be solved by finding the smallest element in the given range K-times and keep updating the upper end of the range to the smallest element found so far. Follow the steps below to solve the problem:

Initialize a global variable, say MinimumElement as INT_MAX.
Declare a function smallestEleUnderRange(root, data) and perform he following operations:
- If root.data is more than data, then update MinimumElement as min of MinimumElement and root.data.
- Iterate over all children of the root. Call recursive function smallestEleUnderRange(child, data).
Declare a function KthSmallestElement(root, k) to perform the following operations:
- Initialize a variable, say ans as INT_MIN, to store the K^th smallest element.
- Iterate over the range [0, K - 1] using a variable i and perform the following:
  - Call smallestEleUnderRange(root, ans) function and then update ans as MinimumElement and then MinimumElement as INT_MAX.
- Finally, print ans as the required answer.

Below is the implementation of the above approach.

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Structure of a node
class Node {
public:
    int data;
    vector<Node*> childs;
};

// Global variable set to Maximum
int MinimumElement = INT_MAX;

// Function that gives the smallest
// element under the range of key
void smallestEleUnderRange(Node* root,
                           int data)
{
    if (root->data > data) {
        MinimumElement = min(
            root->data, MinimumElement);
    }
    for (Node* child : root->childs) {
        smallestEleUnderRange(child, data);
    }
}

// Function to find the Kth smallest element
int kthSmallestElement(Node* root, int k)
{
    int ans = INT_MIN;
    for (int i = 0; i < k; i++) {
        smallestEleUnderRange(root, ans);
        ans = MinimumElement;
        MinimumElement = INT_MAX;
    }
    return ans;
}

// Function to create a new node
Node* newNode(int data)
{
    Node* temp = new Node();
    temp->data = data;
    return temp;
}

// Driver Code
int main()
{
    /*   Let us create below tree
     *              10
     *        /   /    \   \
     *        2  34    56   100
     *       / \         |   /  | \
     *      77  88       1   7  8  9
     */

    Node* root = newNode(10);
    (root->childs).push_back(newNode(2));
    (root->childs).push_back(newNode(34));
    (root->childs).push_back(newNode(56));
    (root->childs).push_back(newNode(100));
    (root->childs[0]->childs).push_back(newNode(77));
    (root->childs[0]->childs).push_back(newNode(88));
    (root->childs[2]->childs).push_back(newNode(1));
    (root->childs[3]->childs).push_back(newNode(7));
    (root->childs[3]->childs).push_back(newNode(8));
    (root->childs[3]->childs).push_back(newNode(9));

    cout << kthSmallestElement(root, 3);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;
public class Main
{
    // Class containing left and
    // right child of current
    // node and key value
    static class Node {
        
        public int data;
        public Vector<Node> childs;
        
        public Node(int data)
        {
            this.data = data;
            childs = new Vector<Node>();
        }
    }
    
    // Global variable set to Maximum
    static int MinimumElement = Integer.MAX_VALUE;
      
    // Function that gives the smallest
    // element under the range of key
    static void smallestEleUnderRange(Node root, int data)
    {
        if (root.data > data) {
            MinimumElement = Math.min(root.data, MinimumElement);
        }
        for(Node child : root.childs) {
            smallestEleUnderRange(child, data);
        }
    }
      
    // Function to find the Kth smallest element
    static int kthSmallestElement(Node root, int k)
    {
        int ans = Integer.MIN_VALUE;
        for (int i = 0; i < k; i++) {
            smallestEleUnderRange(root, ans);
            ans = MinimumElement;
            MinimumElement = Integer.MAX_VALUE;
        }
        return ans;
    }
      
    // Function to create a new node
    static Node newNode(int data)
    {
        Node temp = new Node(data);
        return temp;
    }
    
    public static void main(String[] args) {
        /*   Let us create below tree
         *              10
         *        /   /    \   \
         *        2  34    56   100
         *       / \         |   /  | \
         *      77  88       1   7  8  9
         */
       
        Node root = newNode(10);
        (root.childs).add(newNode(2));
        (root.childs).add(newNode(34));
        (root.childs).add(newNode(56));
        (root.childs).add(newNode(100));
        (root.childs.get(0).childs).add(newNode(77));
        (root.childs.get(0).childs).add(newNode(88));
        (root.childs.get(2).childs).add(newNode(1));
        (root.childs.get(3).childs).add(newNode(7));
        (root.childs.get(3).childs).add(newNode(8));
        (root.childs.get(3).childs).add(newNode(9));
       
        System.out.print(kthSmallestElement(root, 3));
    }
}

// This code is contributed by mukesh07.

Python3

# Python3 program for the above approach
import sys

# Structure of a node
class Node:
    def __init__(self, data):
        self.data = data
        self.childs = []

# Global variable set to Maximum
MinimumElement = sys.maxsize

# Function that gives the smallest
# element under the range of key
def smallestEleUnderRange(root, data):
    global MinimumElement
    if root.data > data:
        MinimumElement = min(root.data, MinimumElement)
    for child in range(len(root.childs)):
        smallestEleUnderRange(root.childs[child], data)

# Function to find the Kth smallest element
def kthSmallestElement(root, k):
    global MinimumElement
    ans = -sys.maxsize
    for i in range(k):
        smallestEleUnderRange(root, ans)
        ans = MinimumElement
        MinimumElement = sys.maxsize
    return ans

# Function to create a new node
def newNode(data):
    temp = Node(data)
    return temp

"""   Let us create below tree
 *              10
 *        /   /    \   \
 *        2  34    56   100
 *       / \         |   /  | \
 *      77  88       1   7  8  9
"""

root = newNode(10)
(root.childs).append(newNode(2))
(root.childs).append(newNode(34))
(root.childs).append(newNode(56))
(root.childs).append(newNode(100))
(root.childs[0].childs).append(newNode(77))
(root.childs[0].childs).append(newNode(88))
(root.childs[2].childs).append(newNode(1))
(root.childs[3].childs).append(newNode(7))
(root.childs[3].childs).append(newNode(8))
(root.childs[3].childs).append(newNode(9))

print(kthSmallestElement(root, 3))

# This code is contributed by divyesh072019.

// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {
    
    // Class containing left and
    // right child of current
    // node and key value
    class Node {
       
        public int data;
        public List<Node> childs;
       
        public Node(int data)
        {
            this.data = data;
            childs = new List<Node>();
        }
    }
    
    // Global variable set to Maximum
    static int MinimumElement = Int32.MaxValue;
     
    // Function that gives the smallest
    // element under the range of key
    static void smallestEleUnderRange(Node root, int data)
    {
        if (root.data > data) {
            MinimumElement = Math.Min(root.data, MinimumElement);
        }
        foreach(Node child in root.childs) {
            smallestEleUnderRange(child, data);
        }
    }
     
    // Function to find the Kth smallest element
    static int kthSmallestElement(Node root, int k)
    {
        int ans = Int32.MinValue;
        for (int i = 0; i < k; i++) {
            smallestEleUnderRange(root, ans);
            ans = MinimumElement;
            MinimumElement = Int32.MaxValue;
        }
        return ans;
    }
     
    // Function to create a new node
    static Node newNode(int data)
    {
        Node temp = new Node(data);
        return temp;
    }
    
  static void Main() {
    /*   Let us create below tree
     *              10
     *        /   /    \   \
     *        2  34    56   100
     *       / \         |   /  | \
     *      77  88       1   7  8  9
     */
  
    Node root = newNode(10);
    (root.childs).Add(newNode(2));
    (root.childs).Add(newNode(34));
    (root.childs).Add(newNode(56));
    (root.childs).Add(newNode(100));
    (root.childs[0].childs).Add(newNode(77));
    (root.childs[0].childs).Add(newNode(88));
    (root.childs[2].childs).Add(newNode(1));
    (root.childs[3].childs).Add(newNode(7));
    (root.childs[3].childs).Add(newNode(8));
    (root.childs[3].childs).Add(newNode(9));
  
    Console.Write(kthSmallestElement(root, 3));
  }
}

// This code is contributed by divyeshrabadiya07.

JavaScript

<script>
    // Javascript program for the above approach
    
    // Structure of a node
    class Node
    {
        constructor(data) {
           this.childs = [];
           this.data = data;
        }
    }
    
    // Global variable set to Maximum
    let MinimumElement = Number.MAX_VALUE;

    // Function that gives the smallest
    // element under the range of key
    function smallestEleUnderRange(root, data)
    {
        if (root.data > data) {
            MinimumElement = Math.min(root.data, MinimumElement);
        }
        for(let child = 0; child < (root.childs).length; child++) {
            smallestEleUnderRange(root.childs[child], data);
        }
    }

    // Function to find the Kth smallest element
    function kthSmallestElement(root, k)
    {
        let ans = Number.MIN_VALUE;
        for (let i = 0; i < k; i++) {
            smallestEleUnderRange(root, ans);
            ans = MinimumElement;
            MinimumElement = Number.MAX_VALUE;
        }
        return ans;
    }

    // Function to create a new node
    function newNode(data)
    {
        let temp = new Node(data);
        return temp;
    }
    
    /*   Let us create below tree
     *              10
     *        /   /    \   \
     *        2  34    56   100
     *       / \         |   /  | \
     *      77  88       1   7  8  9
     */
 
    let root = newNode(10);
    (root.childs).push(newNode(2));
    (root.childs).push(newNode(34));
    (root.childs).push(newNode(56));
    (root.childs).push(newNode(100));
    (root.childs[0].childs).push(newNode(77));
    (root.childs[0].childs).push(newNode(88));
    (root.childs[2].childs).push(newNode(1));
    (root.childs[3].childs).push(newNode(7));
    (root.childs[3].childs).push(newNode(8));
    (root.childs[3].childs).push(newNode(9));
 
    document.write(kthSmallestElement(root, 3));
    
    // This code is contributed by rameshtravel07.
</script>

Output:

Time Complexity: O(N * K) where N is the number of nodes in the given tree.
Auxiliary Space: O(1), but the recursion stack uses a maximum of O(N) space.

shrayansh95

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