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; ` `} ` |

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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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