Dado un árbol binario , la tarea es contar el número de Nodes en el árbol binario dado de modo que la ruta desde la raíz hasta ese Node contenga un Node con un valor mayor o igual que ese Node.
Ejemplos:
Input:
6
/ \
7 4
/ \ / \
3 7 1 2
Output: 5
Explanation:
Root node 6 is considered as its the only node
in the path from root to itself.
Node 4 has the minimum value in it's path 6->4.
Node 1 has the minimum value in it's path 6->4->1.
Node 2 has the minimum value in it's path 6->4->2.
Node 3 has the minimum value in it's path 6->7->3.
Input:
8
/ \
6 5
/ \ / \
6 7 3 9
Output: 5
Explanation:
Root node 8 is considered as its the only node
in the path from root to itself.
Node 6 has the minimum value in it's path 8->6.
Node 6 has the minimum value in it's path 8->6->6.
Node 5 has the minimum value in it's path 8->5.
Node 3 has the minimum value in it's path 8->5->3.
Enfoque: la idea es hacer un recorrido de pedido anticipado en el árbol binario dado. Siga los pasos a continuación para resolver el problema:
- Cree una función para calcular el número de Nodes que satisfacen las condiciones dadas.
- Si el Node actual es NULL, regrese al Node anterior.
- Utilice una variable minNodeVal para almacenar el valor de Node mínimo a lo largo de la ruta desde la raíz hasta el Node actual.
- Si el valor del Node actual es menor o igual que minNodeVal , aumente el recuento final en 1 y actualice el valor de minNodeVal .
- Llame a la función para el hijo izquierdo y derecho del Node actual y repita este proceso para cada Node para obtener el recuento total de los Nodes requeridos.
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 tree node
struct Node {
int key;
Node *left, *right;
};
// Function to create new tree node
Node* newNode(int key)
{
Node* temp = new Node;
temp->key = key;
temp->left = temp->right = NULL;
return temp;
}
// Function to find the total
// number of required nodes
void countReqNodes(Node* root,
int minNodeVal, int& ans)
{
// If current node is null then
// return to the parent node
if (root == NULL)
return;
// Check if current node value is
// less than or equal to minNodeVal
if (root->key <= minNodeVal) {
// Update the value of minNodeVal
minNodeVal = root->key;
// Update the count
ans++;
}
// Go to the left subtree
countReqNodes(root->left,
minNodeVal, ans);
// Go to the right subtree
countReqNodes(root->right,
minNodeVal, ans);
}
// Driver Code
int main()
{
/* Binary Tree creation
8
/ \
/ \
6 5
/ \ / \
/ \ / \
6 7 3 9
*/
Node* root = newNode(8);
root->left = newNode(6);
root->right = newNode(5);
root->left->left = newNode(6);
root->left->right = newNode(7);
root->right->left = newNode(3);
root->right->right = newNode(9);
int ans = 0, minNodeVal = INT_MAX;
// Function Call
countReqNodes(root, minNodeVal, ans);
// Print the result
cout << ans;
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
// Structure of a tree node
static class Node
{
int key;
Node left, right;
};
// Function to create new tree node
static Node newNode(int key)
{
Node temp = new Node();
temp.key = key;
temp.left = temp.right = null;
return temp;
}
static int ans;
// Function to find the total
// number of required nodes
static void countReqNodes(Node root,
int minNodeVal)
{
// If current node is null then
// return to the parent node
if (root == null)
return;
// Check if current node value is
// less than or equal to minNodeVal
if (root.key <= minNodeVal)
{
// Update the value of minNodeVal
minNodeVal = root.key;
// Update the count
ans++;
}
// Go to the left subtree
countReqNodes(root.left,
minNodeVal);
// Go to the right subtree
countReqNodes(root.right,
minNodeVal);
}
// Driver Code
public static void main(String[] args)
{
/* Binary Tree creation
8
/ \
/ \
6 5
/ \ / \
/ \ / \
6 7 3 9
*/
Node root = newNode(8);
root.left = newNode(6);
root.right = newNode(5);
root.left.left = newNode(6);
root.left.right = newNode(7);
root.right.left = newNode(3);
root.right.right = newNode(9);
int minNodeVal = Integer.MAX_VALUE;
ans = 0;
// Function Call
countReqNodes(root, minNodeVal);
// Print the result
System.out.print(ans);
}
}
// This code is contributed by Amit Katiyar
Python3
# Python3 program for the above approach import sys ans = 0 # Class of a tree node class Node: def __init__(self, key): self.key = key self.left = None self.right = None # Function to find the total # number of required nodes def countReqNodes(root, minNodeVal): global ans # If current node is null then # return to the parent node if root == None: return # Check if current node value is # less than or equal to minNodeVal if root.key <= minNodeVal: # Update the value of minNodeVal minNodeVal = root.key # Update the count ans += 1 # Go to the left subtree countReqNodes(root.left, minNodeVal) # Go to the right subtree countReqNodes(root.right, minNodeVal) # Driver Code if __name__ == '__main__': # Binary Tree creation # 8 # / \ # / \ # 6 5 # / \ / \ # / \ / \ # 6 7 3 9 # root = Node(8) root.left = Node(6) root.right = Node(5) root.left.left = Node(6) root.left.right = Node(7) root.right.left = Node(3) root.right.right = Node(9) minNodeVal = sys.maxsize # Function Call countReqNodes(root, minNodeVal) # Print the result print(ans) # This code is contributed by mohit kumar 29
C#
// C# program for the above approach
using System;
class GFG{
// Structure of a tree node
public class Node
{
public int key;
public Node left, right;
};
// Function to create new tree node
static Node newNode(int key)
{
Node temp = new Node();
temp.key = key;
temp.left = temp.right = null;
return temp;
}
static int ans;
// Function to find the total
// number of required nodes
static void countReqNodes(Node root,
int minNodeVal)
{
// If current node is null then
// return to the parent node
if (root == null)
return;
// Check if current node value is
// less than or equal to minNodeVal
if (root.key <= minNodeVal)
{
// Update the value of minNodeVal
minNodeVal = root.key;
// Update the count
ans++;
}
// Go to the left subtree
countReqNodes(root.left,
minNodeVal);
// Go to the right subtree
countReqNodes(root.right,
minNodeVal);
}
// Driver Code
public static void Main(String[] args)
{
/* Binary Tree creation
8
/ \
/ \
6 5
/ \ / \
/ \ / \
6 7 3 9
*/
Node root = newNode(8);
root.left = newNode(6);
root.right = newNode(5);
root.left.left = newNode(6);
root.left.right = newNode(7);
root.right.left = newNode(3);
root.right.right = newNode(9);
int minNodeVal = int.MaxValue;
ans = 0;
// Function Call
countReqNodes(root, minNodeVal);
// Print the result
Console.Write(ans);
}
}
// This code is contributed by Amit Katiyar
Javascript
<script>
// Javascript program for the above approach
// Structure of tree node
class Node
{
constructor(key)
{
this.left = null;
this.right = null;
this.key = key;
}
}
// Function to create new tree node
function newNode(key)
{
let temp = new Node(key);
return temp;
}
let ans;
// Function to find the total
// number of required nodes
function countReqNodes(root, minNodeVal)
{
// If current node is null then
// return to the parent node
if (root == null)
return;
// Check if current node value is
// less than or equal to minNodeVal
if (root.key <= minNodeVal)
{
// Update the value of minNodeVal
minNodeVal = root.key;
// Update the count
ans++;
}
// Go to the left subtree
countReqNodes(root.left,
minNodeVal);
// Go to the right subtree
countReqNodes(root.right,
minNodeVal);
}
// Driver code
/* Binary Tree creation
8
/ \
/ \
6 5
/ \ / \
/ \ / \
6 7 3 9
*/
let root = newNode(8);
root.left = newNode(6);
root.right = newNode(5);
root.left.left = newNode(6);
root.left.right = newNode(7);
root.right.left = newNode(3);
root.right.right = newNode(9);
let minNodeVal = Number.MAX_VALUE;
ans = 0;
// Function Call
countReqNodes(root, minNodeVal);
// Print the result
document.write(ans);
// This code is contributed by suresh07
</script>
Producción:
5
Complejidad temporal: O(N)
Espacio auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por kundudinesh007 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA