Immediate Smaller element in an N-ary Tree

Given an element x, task is to find the value of its immediate smaller element.

Example :

Input : x = 30 (for above tree)
Output : Immediate smaller element is 25

Explanation : Elements 2, 15, 20 and 25 are smaller than x i.e, 30, but 25 is the immediate smaller element and hence the answer.
Approach :

  • Let res be the resultant node.
  • Initialize the resultant Node as NULL.
  • For every Node, check if data of root is greater than res, but less than x. if yes, update res.
  • Recursively do the same for all nodes of the given Generic Tree.
  • Return res, and res->key would be the immediate smaller element.

Below is the implementation of above approach :

// C++ program to find immediate Smaller
// Element of a given element in a n-ary tree.
#include <bits/stdc++.h>
using namespace std;

// class of a node of an n-ary tree
class Node {

    int key;
    vector<Node*> child;

    // constructor
    Node(int data)
        key = data;

// Function to find immediate Smaller Element
// of a given number x
void immediateSmallerElementUtil(Node* root, 
                            int x, Node** res)
    if (root == NULL)

    // if root is greater than res, but less
    // than x, then update res
    if (root->key < x)
        if (!(*res) || (*res)->key < root->key)
            *res = root; // Updating res

    // Number of children of root
    int numChildren = root->child.size();

    // Recursive calling for every child
    for (int i = 0; i < numChildren; i++)
        immediateSmallerElementUtil(root->child[i], x, res);


// Function to return immediate Smaller
// Element of x in tree
Node* immediateSmallerElement(Node* root, int x)
    // resultant node
    Node* res = NULL;

    // calling helper function and using
    // pass by reference
    immediateSmallerElementUtil(root, x, &res);

    return res;

// Driver program
int main()
    // Creating a generic tree
    Node* root = new Node(20);
    (root->child).push_back(new Node(2));
    (root->child).push_back(new Node(34));
    (root->child).push_back(new Node(50));
    (root->child).push_back(new Node(60));
    (root->child).push_back(new Node(70));
    (root->child[0]->child).push_back(new Node(15));
    (root->child[0]->child).push_back(new Node(20));
    (root->child[1]->child).push_back(new Node(30));
    (root->child[2]->child).push_back(new Node(40));
    (root->child[2]->child).push_back(new Node(100));
    (root->child[2]->child).push_back(new Node(20));
    (root->child[0]->child[1]->child).push_back(new Node(25));
    (root->child[0]->child[1]->child).push_back(new Node(50));

    int x = 30;

    cout << "Immediate smaller element of " << x << " is ";
    cout << immediateSmallerElement(root, x)->key << endl;

    return 0;

Output :

Immediate smaller element of 30 is 25

Time Complexity : O(N), where N is the number of nodes in N-ary Tree.
Auxiliary Space : O(N), for recursive call(worst case when a node has N number of childs)

In love with a semicolon because sometimes i miss it so badly)

If you like GeeksforGeeks and would like to contribute, you can also write an article using or mail your article to See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Recommended Posts:

3.6 Average Difficulty : 3.6/5.0
Based on 3 vote(s)