# Next Larger element in n-ary tree

• Difficulty Level : Medium
• Last Updated : 11 Feb, 2023

Given a generic tree and a integer x. Find and return the node with next larger element in the tree i.e. find a node just greater than x. Return NULL if no node is present with value greater than x.

For example, in the given tree

x = 10, just greater node value is 12

The idea is maintain a node pointer res, which will contain the final resultant node.
Traverse the tree and check if root data is greater than x. If so, then compare the root data with res data.
If root data is greater than n and less than res data update res.

Implementation:

## C++

 `// CPP program to find next larger element``// in an n-ary tree.``#include ``using` `namespace` `std;` `// Structure of a node of an n-ary tree``struct` `Node {``    ``int` `key;``    ``vector child;``};` `// Utility function to create a new tree node``Node* newNode(``int` `key)``{``    ``Node* temp = ``new` `Node;``    ``temp->key = key;``    ``return` `temp;``}` `void` `nextLargerElementUtil(Node* root, ``int` `x, Node** res)``{``    ``if` `(root == NULL)``        ``return``;` `    ``// if root is less than res but greater than``    ``// x update res``    ``if` `(root->key > x)``        ``if` `(!(*res) || (*res)->key > root->key)``            ``*res = root;   ` `    ``// Number of children of root``    ``int` `numChildren = root->child.size();` `    ``// Recur calling for every child``    ``for` `(``int` `i = 0; i < numChildren; i++)``        ``nextLargerElementUtil(root->child[i], x, res);` `    ``return``;``}` `// Function to find next Greater element of x in tree``Node* nextLargerElement(Node* root, ``int` `x)``{``    ``// resultant node``    ``Node* res = NULL;` `    ``// calling helper function``    ``nextLargerElementUtil(root, x, &res);` `    ``return` `res;``}` `// Driver program``int` `main()``{``    ``/*   Let us create below tree``   ``*             5``   ``*         /   |  \``   ``*         1   2   3``   ``*        /   / \   \``   ``*       15  4   5   6``   ``*/` `    ``Node* root = newNode(5);``    ``(root->child).push_back(newNode(1));``    ``(root->child).push_back(newNode(2));``    ``(root->child).push_back(newNode(3));``    ``(root->child[0]->child).push_back(newNode(15));``    ``(root->child[1]->child).push_back(newNode(4));``    ``(root->child[1]->child).push_back(newNode(5));``    ``(root->child[2]->child).push_back(newNode(6));` `    ``int` `x = 5;` `    ``cout << ``"Next larger element of "` `<< x << ``" is "``;``    ``cout << nextLargerElement(root, x)->key << endl;` `    ``return` `0;``}`

## Java

 `// Java program to find next larger element``// in an n-ary tree.``import` `java.util.*;` `class` `GFG``{` `// Structure of a node of an n-ary tree``static` `class` `Node``{``    ``int` `key;``    ``Vector child;``};``static` `Node res;` `// Utility function to create a new tree node``static` `Node newNode(``int` `key)``{``    ``Node temp = ``new` `Node();``    ``temp.key = key;``    ``temp.child = ``new` `Vector<>();``    ``return` `temp;``}` `static` `void` `nextLargerElementUtil(Node root, ``int` `x)``{``    ``if` `(root == ``null``)``        ``return``;` `    ``// if root is less than res but``    ``// greater than x, update res``    ``if` `(root.key > x)``        ``if` `((res == ``null` `|| (res).key > root.key))``            ``res = root;` `    ``// Number of children of root``    ``int` `numChildren = root.child.size();` `    ``// Recur calling for every child``    ``for` `(``int` `i = ``0``; i < numChildren; i++)``        ``nextLargerElementUtil(root.child.get(i), x);` `    ``return``;``}` `// Function to find next Greater element``// of x in tree``static` `Node nextLargerElement(Node root, ``int` `x)``{``    ``// resultant node``    ``res = ``null``;` `    ``// calling helper function``    ``nextLargerElementUtil(root, x);` `    ``return` `res;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``/* Let us create below tree``    ``*             5``    ``*         / | \``    ``*         1 2 3``    ``*     / / \ \``    ``*     15 4 5 6``    ``*/``    ``Node root = newNode(``5``);``    ``(root.child).add(newNode(``1``));``    ``(root.child).add(newNode(``2``));``    ``(root.child).add(newNode(``3``));``    ``(root.child.get(``0``).child).add(newNode(``15``));``    ``(root.child.get(``1``).child).add(newNode(``4``));``    ``(root.child.get(``1``).child).add(newNode(``5``));``    ``(root.child.get(``2``).child).add(newNode(``6``));` `    ``int` `x = ``5``;` `    ``System.out.print(``"Next larger element of "` `+``                                    ``x + ``" is "``);``    ``System.out.print(nextLargerElement(root, x).key + ``"\n"``);` `}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python program to find next larger element``# in an n-ary tree.``class` `Node:`` ` `# Structure of a node of an n-ary tree``    ``def` `__init__(``self``):``        ``self``.key ``=` `0``        ``self``.child ``=` `[]` `# Utility function to create a new tree node``def` `newNode(key):``    ``temp ``=` `Node()``    ``temp.key ``=` `key``    ``temp.child ``=` `[]``    ``return` `temp` `res ``=` `None``;   ``def` `nextLargerElementUtil(root,x):``    ``global` `res``    ``if` `(root ``=``=` `None``):``        ``return``;``    ` `     ``# if root is less than res but``    ``# greater than x, update res``    ``if` `(root.key > x):``        ``if` `((res ``=``=` `None` `or` `(res).key > root.key)):``            ``res ``=` `root;``            ` `    ``# Number of children of root``    ``numChildren ``=` `len``(root.child)``    ` `     ``# Recur calling for every child``    ``for` `i ``in` `range``(numChildren):``        ``nextLargerElementUtil(root.child[i], x)``    ``return` `  ``# Function to find next Greater element``# of x in tree``def` `nextLargerElement(root,x):``  ` `   ``# resultant node``    ``global` `res``    ``res``=``None``    ` `    ``# Calling helper function``    ``nextLargerElementUtil(root, x)``    ` `    ``return` `res``    ` `    ``# Driver code``root ``=` `newNode(``5``)``(root.child).append(newNode(``1``))``(root.child).append(newNode(``2``))``(root.child).append(newNode(``3``))``(root.child[``0``].child).append(newNode(``15``))``(root.child[``1``].child).append(newNode(``4``))``(root.child[``1``].child).append(newNode(``5``))``(root.child[``2``].child).append(newNode(``6``))` `x ``=` `5``print``(``"Next larger element of "` `, x , ``" is "``,end``=``'')``print``(nextLargerElement(root, x).key)` `# This code is contributed by rag2127.`

## C#

 `// C# program to find next larger element``// in an n-ary tree.``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{` `// Structure of a node of an n-ary tree``class` `Node``{``    ``public` `int` `key;``    ``public` `List child;``};``static` `Node res;` `// Utility function to create a new tree node``static` `Node newNode(``int` `key)``{``    ``Node temp = ``new` `Node();``    ``temp.key = key;``    ``temp.child = ``new` `List();``    ``return` `temp;``}` `static` `void` `nextLargerElementUtil(Node root,``                                  ``int` `x)``{``    ``if` `(root == ``null``)``        ``return``;` `    ``// if root is less than res but``    ``// greater than x, update res``    ``if` `(root.key > x)``        ``if` `((res == ``null` `||``            ``(res).key > root.key))``            ``res = root;` `    ``// Number of children of root``    ``int` `numChildren = root.child.Count;` `    ``// Recur calling for every child``    ``for` `(``int` `i = 0; i < numChildren; i++)``        ``nextLargerElementUtil(root.child[i], x);` `    ``return``;``}` `// Function to find next Greater element``// of x in tree``static` `Node nextLargerElement(Node root,   ``                                  ``int` `x)``{``    ``// resultant node``    ``res = ``null``;` `    ``// calling helper function``    ``nextLargerElementUtil(root, x);` `    ``return` `res;``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``/* Let us create below tree``    ``*             5``    ``*         / | \``    ``*         1 2 3``    ``*     / / \ \``    ``*     15 4 5 6``    ``*/``    ``Node root = newNode(5);``    ``(root.child).Add(newNode(1));``    ``(root.child).Add(newNode(2));``    ``(root.child).Add(newNode(3));``    ``(root.child[0].child).Add(newNode(15));``    ``(root.child[1].child).Add(newNode(4));``    ``(root.child[1].child).Add(newNode(5));``    ``(root.child[2].child).Add(newNode(6));` `    ``int` `x = 5;` `    ``Console.Write(``"Next larger element of "` `+``                                 ``x + ``" is "``);``    ``Console.Write(nextLargerElement(root, x).key + ``"\n"``);``}``}` `// This code is contributed by PrinciRaj1992`

## Javascript

 ``

Output

`Next larger element of 5 is 6`

### Algorithmic steps:

Traverse the tree in post-order and store the nodes in a stack.
Create an empty hash map to store the next larger element for each node.
For each node in the stack, pop it from the stack and find the next larger element for it by checking the elements on the top of the stack. If an element on the top of the stack is greater than the current node, it is the next larger element for the current node. Keep popping elements from the stack until the top of the stack is less than or equal to the current node.
Add the next larger element for the current node to the hash map.
Repeat steps 3 and 4 until the stack is empty.
Return the next larger element for the target node by looking it up in the hash map.

Note: This algorithm assumes that the tree is a binary search tree, where the value of each node is greater than the values of its left child and less than the values of its right child. If the tree is not a binary search tree, the above algorithm may not give correct result

## C++

 `#include ``#include ``#include ` `struct` `TreeNode {``    ``int` `val;``    ``vector children;``    ``TreeNode(``int` `x) : val(x) {}``};` `void` `postOrder(TreeNode *root, stack<``int``> &nodes) {``    ``if` `(!root) ``return``;``    ``for` `(``int` `i = 0; i < root->children.size(); i++) {``        ``postOrder(root->children[i], nodes);``    ``}``    ``nodes.push(root->val);``}` `unordered_map<``int``, ``int``> findNextLarger(TreeNode *root) {``    ``stack<``int``> nodes;``    ``unordered_map<``int``, ``int``> nextLarger;``    ``postOrder(root, nodes);` `    ``while` `(!nodes.empty()) {``        ``int` `current = nodes.top();``        ``nodes.pop();``        ``while` `(!nodes.empty() && nodes.top() <= current) {``            ``nodes.pop();``        ``}``        ``if` `(!nodes.empty()) {``            ``nextLarger[current] = nodes.top();``        ``} ``else` `{``            ``nextLarger[current] = -1;``        ``}``    ``}` `    ``return` `nextLarger;``}` `int` `main() {``    ``TreeNode *root = ``new` `TreeNode(8);``    ``root->children.push_back(``new` `TreeNode(3));``    ``root->children.push_back(``new` `TreeNode(10));``    ``root->children[0]->children.push_back(``new` `TreeNode(1));``    ``root->children[0]->children.push_back(``new` `TreeNode(6));``    ``root->children[0]->children[1]->children.push_back(``new` `TreeNode(4));``    ``root->children[0]->children[1]->children.push_back(``new` `TreeNode(7));``    ``root->children[1]->children.push_back(``new` `TreeNode(14));` `    ``int` `target = 4;``    ``unordered_map<``int``, ``int``> nextLarger = findNextLarger(root);``    ``cout << ``"The next larger element for "` `<< target << ``" is: "` `<< nextLarger[target] << endl;``    ``return` `0;``}`

`The next larger element for 4 is 7`

#### Time and Space complexities are:

The time complexity of this algorithm is O(n), where n is the number of nodes in the n-ary tree. The reason for this is that we traverse each node in the tree once, and for each node, we perform a constant amount of work to find its next larger element.

The auxiliary space of this algorithm is O(n), where n is the number of nodes in the n-ary tree. The reason for this is that we use a stack to store the nodes in post-order and a hash map to store the next larger element for each node. The size of the stack and the hash map is proportional to the number of nodes in the tree, so their combined space complexity is O(n).

This article is contributed by Chhavi. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

My Personal Notes arrow_drop_up