# 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
be the resultant node.**res** - 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 {` ` ` `public` `:` ` ` `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)` ` ` `return` `;` ` ` ` ` `// 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);` ` ` ` ` `return` `;` `}` ` ` `// 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)

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