Dado un árbol N-ario que consta de N Nodes, la tarea es encontrar el Node que tiene el mayor valor en el árbol N -ario dado .
Ejemplos:
Aporte:
Salida: 90
Explicación: El Node con el mayor valor en el árbol es 90.Aporte:
Salida: 95
Explicación: El Node con el mayor valor en el árbol es 95.
Enfoque: El problema dado se puede resolver atravesando el árbol N -ario dado y haciendo un seguimiento del valor máximo de los Nodes que ocurrieron. Después de completar el recorrido, imprima el valor máximo obtenido.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Structure of a
// node of N-ary tree
struct Node {
int key;
vector<Node*> child;
};
// Stores the node with largest value
Node* maximum = NULL;
// Function to create a new Node
Node* newNode(int key)
{
Node* temp = new Node;
temp->key = key;
// Return the newly created node
return temp;
}
// Function to find the node with
// largest value in N-ary tree
void findlargest(Node* root)
{
// Base Case
if (root == NULL)
return;
// If maximum is NULL, return
// the value of root node
if ((maximum) == NULL)
maximum = root;
// If value of the root is greater
// than maximum, update the maximum node
else if (root->key > (maximum)->key) {
maximum = root;
}
// Recursively call for all the
// children of the root node
for (int i = 0;
i < root->child.size(); i++) {
findlargest(root->child[i]);
}
}
// Driver Code
int main()
{
// Given N-ary tree
Node* root = newNode(11);
(root->child).push_back(newNode(21));
(root->child).push_back(newNode(29));
(root->child).push_back(newNode(90));
(root->child[0]->child).push_back(newNode(18));
(root->child[1]->child).push_back(newNode(10));
(root->child[1]->child).push_back(newNode(12));
(root->child[2]->child).push_back(newNode(77));
findlargest(root);
// Print the largest value
cout << maximum->key;
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
// Structure of a
// node of N-ary tree
static class Node
{
int key;
Vector<Node> child = new Vector<>();
};
// Stores the node with largest value
static Node maximum = null;
// Function to create a new Node
static Node newNode(int key)
{
Node temp = new Node();
temp.key = key;
// Return the newly created node
return temp;
}
// Function to find the node with
// largest value in N-ary tree
static void findlargest(Node root)
{
// Base Case
if (root == null)
return;
// If maximum is null, return
// the value of root node
if ((maximum) == null)
maximum = root;
// If value of the root is greater
// than maximum, update the maximum node
else if (root.key > (maximum).key)
{
maximum = root;
}
// Recursively call for all the
// children of the root node
for(int i = 0;
i < root.child.size(); i++)
{
findlargest(root.child.get(i));
}
}
// Driver Code
public static void main(String[] args)
{
// Given N-ary tree
Node root = newNode(11);
(root.child).add(newNode(21));
(root.child).add(newNode(29));
(root.child).add(newNode(90));
(root.child.get(0).child).add(newNode(18));
(root.child.get(1).child).add(newNode(10));
(root.child.get(1).child).add(newNode(12));
(root.child.get(2).child).add(newNode(77));
findlargest(root);
// Print the largest value
System.out.print(maximum.key);
}
}
// This code is contributed by Princi Singh
Python3
# Python3 program for the above approach # Structure of a # node of N-ary tree class Node: # Constructor to set the data of # the newly created tree node def __init__(self, key): self.key = key self.child = [] # Stores the node with largest value maximum = None # Function to create a new Node def newNode(key): temp = Node(key) # Return the newly created node return temp # Function to find the node with # largest value in N-ary tree def findlargest(root): global maximum # Base Case if (root == None): return # If maximum is null, return # the value of root node if ((maximum) == None): maximum = root # If value of the root is greater # than maximum, update the maximum node elif (root.key > (maximum).key): maximum = root # Recursively call for all the # children of the root node for i in range(len(root.child)): findlargest(root.child[i]) # Given N-ary tree root = newNode(11) (root.child).append(newNode(21)) (root.child).append(newNode(29)) (root.child).append(newNode(90)) (root.child[0].child).append(newNode(18)) (root.child[1].child).append(newNode(10)) (root.child[1].child).append(newNode(12)) (root.child[2].child).append(newNode(77)) findlargest(root) # Print the largest value print(maximum.key) # This code is contributed by decode2207.
C#
// C# program for the above approach
using System;
using System.Collections.Generic;
public class GFG{
// Structure of a
// node of N-ary tree
class Node
{
public int key;
public List<Node> child = new List<Node>();
};
// Stores the node with largest value
static Node maximum = null;
// Function to create a new Node
static Node newNode(int key)
{
Node temp = new Node();
temp.key = key;
// Return the newly created node
return temp;
}
// Function to find the node with
// largest value in N-ary tree
static void findlargest(Node root)
{
// Base Case
if (root == null)
return;
// If maximum is null, return
// the value of root node
if ((maximum) == null)
maximum = root;
// If value of the root is greater
// than maximum, update the maximum node
else if (root.key > (maximum).key)
{
maximum = root;
}
// Recursively call for all the
// children of the root node
for(int i = 0;
i < root.child.Count; i++)
{
findlargest(root.child[i]);
}
}
// Driver Code
public static void Main(String[] args)
{
// Given N-ary tree
Node root = newNode(11);
(root.child).Add(newNode(21));
(root.child).Add(newNode(29));
(root.child).Add(newNode(90));
(root.child[0].child).Add(newNode(18));
(root.child[1].child).Add(newNode(10));
(root.child[1].child).Add(newNode(12));
(root.child[2].child).Add(newNode(77));
findlargest(root);
// Print the largest value
Console.Write(maximum.key);
}
}
// This code is contributed by 29AjayKumar
Javascript
<script>
// Javascript program for the above approach
// Structure of a
// node of N-ary tree
class Node
{
constructor(key) {
this.key = key;
this.child = [];
}
}
// Stores the node with largest value
let maximum = null;
// Function to create a new Node
function newNode(key)
{
let temp = new Node(key);
// Return the newly created node
return temp;
}
// Function to find the node with
// largest value in N-ary tree
function findlargest(root)
{
// Base Case
if (root == null)
return;
// If maximum is null, return
// the value of root node
if ((maximum) == null)
maximum = root;
// If value of the root is greater
// than maximum, update the maximum node
else if (root.key > (maximum).key)
{
maximum = root;
}
// Recursively call for all the
// children of the root node
for(let i = 0; i < root.child.length; i++)
{
findlargest(root.child[i]);
}
}
// Given N-ary tree
let root = newNode(11);
(root.child).push(newNode(21));
(root.child).push(newNode(29));
(root.child).push(newNode(90));
(root.child[0].child).push(newNode(18));
(root.child[1].child).push(newNode(10));
(root.child[1].child).push(newNode(12));
(root.child[2].child).push(newNode(77));
findlargest(root);
// Print the largest value
document.write(maximum.key);
// This code is contributed by surehs07.
</script>
Producción:
90
Complejidad temporal: O(N)
Espacio auxiliar: O(1)