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Immediate Smaller element in an N-ary Tree

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  • Difficulty Level : Medium
  • Last Updated : 12 Aug, 2022
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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;


Immediate smaller element of 30 is 25

Complexity Analysis:

  • 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)

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