Next Larger element in n-ary tree
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++
#include <bits/stdc++.h>
using namespace std;
struct Node {
int key;
vector<Node*> child;
};
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->key > x)
if (!(*res) || (*res)->key > root->key)
*res = root;
int numChildren = root->child.size();
for ( int i = 0; i < numChildren; i++)
nextLargerElementUtil(root->child[i], x, res);
return ;
}
Node* nextLargerElement(Node* root, int x)
{
Node* res = NULL;
nextLargerElementUtil(root, x, &res);
return res;
}
int main()
{
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
import java.util.*;
class GFG
{
static class Node
{
int key;
Vector<Node> child;
};
static Node res;
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.key > x)
if ((res == null || (res).key > root.key))
res = root;
int numChildren = root.child.size();
for ( int i = 0 ; i < numChildren; i++)
nextLargerElementUtil(root.child.get(i), x);
return ;
}
static Node nextLargerElement(Node root, int x)
{
res = null ;
nextLargerElementUtil(root, x);
return res;
}
public static void main(String[] args)
{
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" );
}
}
|
Python3
class Node:
def __init__( self ):
self .key = 0
self .child = []
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.key > x):
if ((res = = None or (res).key > root.key)):
res = root;
numChildren = len (root.child)
for i in range (numChildren):
nextLargerElementUtil(root.child[i], x)
return
def nextLargerElement(root,x):
global res
res = None
nextLargerElementUtil(root, x)
return res
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)
|
C#
using System;
using System.Collections.Generic;
class GFG
{
class Node
{
public int key;
public List<Node> child;
};
static Node res;
static Node newNode( int key)
{
Node temp = new Node();
temp.key = key;
temp.child = new List<Node>();
return temp;
}
static void nextLargerElementUtil(Node root,
int x)
{
if (root == null )
return ;
if (root.key > x)
if ((res == null ||
(res).key > root.key))
res = root;
int numChildren = root.child.Count;
for ( int i = 0; i < numChildren; i++)
nextLargerElementUtil(root.child[i], x);
return ;
}
static Node nextLargerElement(Node root,
int x)
{
res = null ;
nextLargerElementUtil(root, x);
return res;
}
public static void Main(String[] args)
{
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" );
}
}
|
Javascript
<script>
class Node
{
constructor()
{
this .key = 0;
this .child = [];
}
};
var res = null ;
function newNode(key)
{
var temp = new Node();
temp.key = key;
temp.child = [];
return temp;
}
function nextLargerElementUtil(root, x)
{
if (root == null )
return ;
if (root.key > x)
if ((res == null ||
(res).key > root.key))
res = root;
var numChildren = root.child.length;
for ( var i = 0; i < numChildren; i++)
nextLargerElementUtil(root.child[i], x);
return ;
}
function nextLargerElement(root, x)
{
res = null ;
nextLargerElementUtil(root, x);
return res;
}
var root = newNode(5);
(root.child).push(newNode(1));
(root.child).push(newNode(2));
(root.child).push(newNode(3));
(root.child[0].child).push(newNode(15));
(root.child[1].child).push(newNode(4));
(root.child[1].child).push(newNode(5));
(root.child[2].child).push(newNode(6));
var x = 5;
document.write( "Next larger element of " +
x + " is " );
document.write(nextLargerElement(root, x).key + "<br>" );
</script>
|
Output
Next larger element of 5 is 6
Time complexity :- O(N)
Space complexity :- O(H)
Example no2:
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
Program: Implementing by Postorder traversing tree:
C++
#include <iostream>
#include <stack>
#include <unordered_map>
struct TreeNode {
int val;
vector<TreeNode*> 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;
}
|
Java
import java.util.*;
class TreeNode {
int val;
List<TreeNode> children;
TreeNode( int x) {
val = x;
children = new ArrayList<>();
}
}
public class Main {
static void postOrder(TreeNode root, Stack<Integer> nodes) {
if (root == null )
return ;
for (TreeNode child : root.children) {
postOrder(child, nodes);
}
nodes.push(root.val);
}
static Map<Integer, Integer> findNextLarger(TreeNode root) {
Stack<Integer> nodes = new Stack<>();
Map<Integer, Integer> nextLarger = new HashMap<>();
postOrder(root, nodes);
Stack<Integer> stack = new Stack<>();
for ( int i = nodes.size() - 1 ; i >= 0 ; i--) {
while (!stack.isEmpty() && stack.peek() <= nodes.get(i)) {
stack.pop();
}
nextLarger.put(nodes.get(i), stack.isEmpty() ? - 1 : stack.peek());
stack.push(nodes.get(i));
}
return nextLarger;
}
public static void main(String[] args) {
TreeNode root = new TreeNode( 8 );
root.children = new ArrayList<>();
root.children.add( new TreeNode( 3 ));
root.children.add( new TreeNode( 10 ));
root.children.get( 0 ).children = new ArrayList<>();
root.children.get( 0 ).children.add( new TreeNode( 1 ));
root.children.get( 0 ).children.add( new TreeNode( 6 ));
root.children.get( 0 ).children.get( 1 ).children = new ArrayList<>();
root.children.get( 0 ).children.get( 1 ).children.add( new TreeNode( 4 ));
root.children.get( 0 ).children.get( 1 ).children.add( new TreeNode( 7 ));
root.children.get( 1 ).children = new ArrayList<>();
root.children.get( 1 ).children.add( new TreeNode( 14 ));
int target = 4 ;
Map<Integer, Integer> nextLarger = findNextLarger(root);
Integer nextLargerValue = nextLarger.get(target);
System.out.println( "The next larger element for " + target + " is: " + (nextLargerValue != null ? nextLargerValue : - 1 ));
}
}
|
Python3
class Node:
def __init__( self ):
self .vcal = 0
self .children = []
def newNode(val):
temp = Node()
temp.val = val
temp.children = []
return temp
def postOrder(root, nodes):
if not root:
return
for child in root.children:
postOrder(child, nodes)
nodes.append(root.val)
def findNextLarger(root):
nodes = []
nextLarger = {}
postOrder(root, nodes)
stack = []
for num in nodes[:: - 1 ]:
while stack and stack[ - 1 ] < = num:
stack.pop()
nextLarger[num] = stack[ - 1 ] if stack else - 1
stack.append(num)
return nextLarger
root = newNode( 8 )
root.children.append(newNode( 3 ))
root.children.append(newNode( 10 ))
root.children[ 0 ].children.append(newNode( 1 ))
root.children[ 0 ].children.append(newNode( 6 ))
root.children[ 0 ].children[ 1 ].children.append(newNode( 4 ))
root.children[ 0 ].children[ 1 ].children.append(newNode( 7 ))
root.children[ 1 ].children.append(newNode( 14 ))
target = 4
nextLarger = findNextLarger(root)
print (f "The next larger element for {target} is: {nextLarger[target]}" )
|
C#
using System;
using System.Collections.Generic;
class Node {
public int val;
public List<Node> children;
public Node()
{
val = 0;
children = new List<Node>();
}
}
public class MainClass {
static Node NewNode( int val)
{
Node temp = new Node();
temp.val = val;
temp.children = new List<Node>();
return temp;
}
static void PostOrder(Node root, List< int > nodes)
{
if (root == null )
return ;
foreach (Node child in root.children)
{
PostOrder(child, nodes);
}
nodes.Add(root.val);
}
static Dictionary< int , int > FindNextLarger(Node root)
{
List< int > nodes = new List< int >();
Dictionary< int , int > nextLarger
= new Dictionary< int , int >();
PostOrder(root, nodes);
Stack< int > stack = new Stack< int >();
for ( int i = nodes.Count - 1; i >= 0; i--) {
while (stack.Count > 0
&& stack.Peek() <= nodes[i]) {
stack.Pop();
}
if (stack.Count > 0) {
nextLarger[nodes[i]] = stack.Peek();
}
else {
nextLarger[nodes[i]] = -1;
}
stack.Push(nodes[i]);
}
return nextLarger;
}
public static void Main()
{
Node root = NewNode(8);
root.children.Add(NewNode(3));
root.children.Add(NewNode(10));
root.children[0].children.Add(NewNode(1));
root.children[0].children.Add(NewNode(6));
root.children[0].children[1].children.Add(
NewNode(4));
root.children[0].children[1].children.Add(
NewNode(7));
root.children[1].children.Add(NewNode(14));
int target = 4;
Dictionary< int , int > nextLarger
= FindNextLarger(root);
int nextLargerValue;
if (nextLarger.TryGetValue(target,
out nextLargerValue)) {
Console.WriteLine( "The next larger element for "
+ target
+ " is: " + nextLargerValue);
}
else {
Console.WriteLine( "The next larger element for "
+ target + " is: -1" );
}
}
}
|
Javascript
class TreeNode {
constructor(x) {
this .val = x;
this .children = [];
}
}
function postOrder(root, nodes) {
if (!root) return ;
for (const child of root.children) {
postOrder(child, nodes);
}
nodes.push(root.val);
}
function findNextLarger(root) {
const nodes = [];
const nextLarger = new Map();
postOrder(root, nodes);
const stack = [];
for (let i = nodes.length - 1; i >= 0; i--) {
while (stack.length > 0 && stack[stack.length - 1] <= nodes[i]) {
stack.pop();
}
nextLarger.set(nodes[i], stack.length > 0 ? stack[stack.length - 1] : -1);
stack.push(nodes[i]);
}
return nextLarger;
}
const root = new TreeNode(8);
root.children.push( new TreeNode(3));
root.children.push( new TreeNode(10));
root.children[0].children.push( new TreeNode(1));
root.children[0].children.push( new TreeNode(6));
root.children[0].children[1].children.push( new TreeNode(4));
root.children[0].children[1].children.push( new TreeNode(7));
root.children[1].children.push( new TreeNode(14));
const target = 4;
const nextLarger = findNextLarger(root);
const nextLargerValue = nextLarger.get(target);
console.log(`The next larger element for ${target} is: ${nextLargerValue !== undefined ? nextLargerValue : -1}`);
|
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).
Last Updated :
16 Aug, 2023
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