Dado un árbol binario, la tarea es encontrar el número de Nodes visibles en el árbol binario dado. Un Node es un Node visible si, en el camino desde la raíz hasta el Node N, no hay ningún Node con valor mayor que N,
Ejemplos:
Input:
5
/ \
3 10
/ \ /
20 21 1
Output: 4
Explanation:
There are 4 visible nodes.
They are:
5: In the path 5 -> 3, 5 is the highest node value.
20: In the path 5 -> 3 -> 20, 20 is the highest node value.
21: In the path 5 -> 3 -> 21, 21 is the highest node value.
10: In the path 5 -> 10 -> 1, 10 is the highest node value.
Input:
-1
\
-2
\
-3
Output: 1
Aproximación: La idea es atravesar primero el árbol. Dado que necesitamos ver el valor máximo en la ruta dada, el recorrido de pedido previo se usa para atravesar el árbol binario dado. Mientras recorremos el árbol, debemos mantener un registro del valor máximo del Node que hemos visto hasta ahora. Si el Node actual es mayor o igual que el valor máximo, incremente el recuento del Node visible y actualice el valor máximo con el valor del Node actual.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation to
// count the number of
// visible nodes in the
// binary tree
#include <bits/stdc++.h>
using namespace std;
// Node containing the
// left and right child of
// current node and the key value
struct Node
{
int data;
Node *left, *right;
};
/* Utility that allocates a
new node with the given key and
NULL left and right pointers. */
struct Node* newnode(int data)
{
struct Node* node = new (struct Node);
node->data = data;
node->left = node->right = NULL;
return (node);
}
// Variable to keep the track
// of visible nodes
int countNode = 0;
// Function to perform the preorder traversal
// for the given tree
void preOrder(Node* node, int mx)
{
// Base case
if (node == NULL)
{
return;
}
// If the current node value is greater
// or equal to the max value,
// then update count variable
// and also update max variable
if (node->data >= mx)
{
countNode++;
mx = max(node->data, mx);
}
// Traverse to the left
preOrder(node->left, mx);
// Traverse to the right
preOrder(node->right, mx);
}
// Driver code
int main()
{
struct Node* root = newnode(5);
/*
5
/ \
3 10
/ \ /
20 21 1
*/
root->left = newnode(3);
root->right = newnode(10);
root->left->left = newnode(20);
root->left->right = newnode(21);
root->right->left = newnode(1);
preOrder(root, INT_MIN);
cout << countNode;
}
// This code is contributed by gauravrajput1
Java
// Java implementation to count the
// number of visible nodes in
// the binary tree
// Class containing the left and right
// child of current node and the
// key value
class Node {
int data;
Node left, right;
// Constructor of the class
public Node(int item)
{
data = item;
left = right = null;
}
}
public class GFG {
Node root;
// Variable to keep the track
// of visible nodes
static int count;
// Function to perform the preorder traversal
// for the given tree
static void preOrder(Node node, int max)
{
// Base case
if (node == null) {
return;
}
// If the current node value is greater
// or equal to the max value,
// then update count variable
// and also update max variable
if (node.data >= max) {
count++;
max = Math.max(node.data, max);
}
// Traverse to the left
preOrder(node.left, max);
// Traverse to the right
preOrder(node.right, max);
}
// Driver code
public static void main(String[] args)
{
GFG tree = new GFG();
/*
5
/ \
3 10
/ \ /
20 21 1
*/
tree.root = new Node(5);
tree.root.left = new Node(3);
tree.root.right = new Node(10);
tree.root.left.left = new Node(20);
tree.root.left.right = new Node(21);
tree.root.right.left = new Node(1);
count = 0;
preOrder(tree.root, Integer.MIN_VALUE);
System.out.println(count);
}
}
Python3
# Python 3 implementation to # count the number of # visible nodes in the # binary tree import sys # Node containing the # left and right child of # current node and the key value class newNode: def __init__(self, data): self.data = data self.left = None self.right = None ''' Utility that allocates a new node with the given key and None left and right pointers. */ ''' # Variable to keep the track # of visible nodes countNode = 0 # Function to perform the # preorder traversal for the # given tree def preOrder(node, mx): global countNode # Base case if (node == None): return # If the current node value # is greater or equal to the # max value, then update count # variable and also update max # variable if (node.data >= mx): countNode += 1 mx = max(node.data, mx) # Traverse to the left preOrder(node.left, mx) # Traverse to the right preOrder(node.right, mx) # Driver code if __name__ == '__main__': root = newNode(5) ''' /* 5 / \ 3 10 / / / 20 21 1 */ ''' root.left = newNode(3) root.right = newNode(10) root.left.left = newNode(20) root.left.right = newNode(21) root.right.left = newNode(1) preOrder(root, -sys.maxsize-1) print(countNode) # This code is contributed by SURENDRA_GANGWAR
C#
// C# implementation to count the
// number of visible nodes in
// the binary tree
using System;
// Class containing the left and right
// child of current node and the
// key value
class Node
{
public int data;
public Node left, right;
// Constructor of the class
public Node(int item)
{
data = item;
left = right = null;
}
}
class GFG{
Node root;
// Variable to keep the track
// of visible nodes
static int count;
// Function to perform the preorder
// traversal for the given tree
static void preOrder(Node node, int max)
{
// Base case
if (node == null)
{
return;
}
// If the current node value is greater
// or equal to the max value,
// then update count variable
// and also update max variable
if (node.data >= max)
{
count++;
max = Math.Max(node.data, max);
}
// Traverse to the left
preOrder(node.left, max);
// Traverse to the right
preOrder(node.right, max);
}
// Driver code
static public void Main(String[] args)
{
GFG tree = new GFG();
/*
5
/ \
3 10
/ \ /
20 21 1
*/
tree.root = new Node(5);
tree.root.left = new Node(3);
tree.root.right = new Node(10);
tree.root.left.left = new Node(20);
tree.root.left.right = new Node(21);
tree.root.right.left = new Node(1);
count = 0;
preOrder(tree.root, int.MinValue);
Console.WriteLine(count);
}
}
// This code is contributed by Amit Katiyar
Javascript
<script>
// Javascript implementation to count the
// number of visible nodes in the binary tree
// Class containing the left and right
// child of current node and the
// key value
class Node
{
constructor(item) {
this.left = null;
this.right = null;
this.data = item;
}
}
let root;
// Variable to keep the track
// of visible nodes
let count;
// Function to perform the preorder traversal
// for the given tree
function preOrder(node, max)
{
// Base case
if (node == null) {
return;
}
// If the current node value is greater
// or equal to the max value,
// then update count variable
// and also update max variable
if (node.data >= max) {
count++;
max = Math.max(node.data, max);
}
// Traverse to the left
preOrder(node.left, max);
// Traverse to the right
preOrder(node.right, max);
}
/*
5
/ \
3 10
/ \ /
20 21 1
*/
root = new Node(5);
root.left = new Node(3);
root.right = new Node(10);
root.left.left = new Node(20);
root.left.right = new Node(21);
root.right.left = new Node(1);
count = 0;
preOrder(root, Number.MIN_VALUE);
document.write(count);
// This code is contributed by rameshtravel07.
</script>
Producción:
4
Análisis de
Complejidad: Complejidad de Tiempo: O(N) donde N es un número de Nodes en el árbol Binario.
Espacio Auxiliar: O(H) donde H es la altura del árbol Binario.