Dado un árbol binario y un número, la devolución es verdadera si el árbol tiene un camino de raíz a hoja tal que el producto de todos los valores a lo largo de ese camino es igual al número dado. El retorno es falso si no se puede encontrar tal ruta.
Por ejemplo, en el árbol anterior, existen tres caminos de raíces a hojas con los siguientes productos.
- 240 –> 10 – 8 – 3
- 400 –> 10 – 8 – 5
- 40 –> 10 – 2 – 2
Enfoque: la idea es comenzar a recorrer el árbol recursivamente y dividir el valor del Node actual del producto si es divisible cuando se repite hacia abajo, y verificar si el producto es 1 cuando llegue al Node hoja de la ruta actual del árbol .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to check if there exists
// a root to leaf path product with
// given product
#include <bits/stdc++.h>
using namespace std;
// Binary Tree Node
struct node {
int data;
struct node* left;
struct node* right;
};
// Function to check if there exists a path
// with given product
// Strategy: divide the node value from the product
// if it is divisible when recurring down, and check
// to see if the product is 1 when you reach leaf
// node of the current path out of tree.
bool hasPathProduct(struct node* node, int prod)
{
// return true if we run out
// of tree and prod==1
if (node == NULL) {
return (prod == 1);
}
else {
bool ans = 1;
// Check if product is divisible by
// current node, if not we are on wrong path
if (prod % (node->data))
return false;
// otherwise check both subtrees
int subProduct = prod / node->data;
// If we reach a leaf node and prod
// becomes 1 then return true
if (subProduct == 1 && node->left == NULL
&& node->right == NULL)
return 1;
if (node->left)
ans = hasPathProduct(node->left, subProduct);
if (node->right)
ans = hasPathProduct(node->right, subProduct);
return ans;
}
}
/* UTILITY FUNCTIONS */
// Helper function that allocates
// a new node with the given data
// and NULL left and right pointers
struct node* newnode(int data)
{
struct node* newNode = new node();
newNode->data = data;
newNode->left = NULL;
newNode->right = NULL;
return (newNode);
}
// Driver Code
int main()
{
int prod = 400;
/* Constructed binary tree is
10
/ \
8 2
/ \ /
3 5 2
*/
struct node* root = newnode(10);
root->left = newnode(8);
root->right = newnode(2);
root->left->left = newnode(3);
root->left->right = newnode(5);
root->right->left = newnode(2);
if (hasPathProduct(root, prod))
cout << "YES";
else
cout << "NO";
return 0;
}
Java
// Java program to check if there exists
// a root to leaf path product with
// given product
class GFG
{
// Binary Tree Node
static class node
{
int data;
node left;
node right;
};
// Function to check if there exists a path
// with given product
// Strategy: divide the node value from the product
// if it is divisible when recurring down, and check
// to see if the product is 1 when you reach leaf
// node of the current path out of tree.
static boolean hasPathProduct(node node, int prod)
{
// return true if we run out
// of tree and prod==1
if (node == null)
{
return (prod == 1);
}
else
{
boolean ans = true;
// Check if product is divisible by
// current node, if not we are on wrong path
if (prod % (node.data) == 1)
{
return false;
}
// otherwise check both subtrees
int subProduct = prod / node.data;
// If we reach a leaf node and prod
// becomes 1 then return true
if (subProduct == 1 && node.left == null
&& node.right == null)
{
return true;
}
if (node.left != null)
{
ans = hasPathProduct(node.left, subProduct);
}
if (node.right != null)
{
ans = hasPathProduct(node.right, subProduct);
}
return ans;
}
}
/* UTILITY FUNCTIONS */
// Helper function that allocates
// a new node with the given data
// and null left and right pointers
static node newnode(int data)
{
node node = new node();
node.data = data;
node.left = null;
node.right = null;
return (node);
}
// Driver Code
public static void main(String[] args)
{
int prod = 400;
/* Constructed binary tree is
10
/ \
8 2
/ \ /
3 5 2
*/
node root = newnode(10);
root.left = newnode(8);
root.right = newnode(2);
root.left.left = newnode(3);
root.left.right = newnode(5);
root.right.left = newnode(2);
if (hasPathProduct(root, prod))
{
System.out.println("Yes");
}
else
{
System.out.println("No");
}
}
}
// This code is contributed by Princi Singh
Python3
# Python3 program to check if there exists
# a root to leaf path product with
# given product
""" UTILITY FUNCTIONS """
# Helper function that allocates
# a new node with the given data
# and None left and right pointers
class newnode:
def __init__(self, data):
self.data = data
self.left = self.right = None
# Function to check if there exists
# a path with given product
# Strategy: divide the node value from
# the product if it is divisible when
# recurring down, and check to see if
# the product is 1 when you reach leaf
# node of the current path out of tree.
def hasPathProduct(node, prod) :
# return true if we run out
# of tree and prod==1
if (node == None) :
return (prod == 1)
else :
ans = 1
# Check if product is divisible by
# current node, if not we are on wrong path
if (prod % (node.data)) :
return False
# otherwise check both subtrees
subProduct = prod // node.data
# If we reach a leaf node and prod
# becomes 1 then return true
if (subProduct == 1 and
node.left == None and
node.right == None) :
return 1
if (node.left) :
ans = hasPathProduct(node.left,
subProduct)
if (node.right) :
ans = hasPathProduct(node.right,
subProduct)
return ans
# Driver Code
if __name__ == '__main__':
prod = 400
""" Constructed binary tree is
10
/ \
8 2
/ \ /
3 5 2
"""
root = newnode(10)
root.left = newnode(8)
root.right = newnode(2)
root.left.left = newnode(3)
root.left.right = newnode(5)
root.right.left = newnode(2)
if (hasPathProduct(root, prod)) :
print("YES" )
else:
print("NO")
# This code is contributed
# by SHUBHAMSINGH10
C#
// C# program to check if there exists
// a root to leaf path product with
// given product
using System;
class GFG
{
// Binary Tree Node
public class node
{
public int data;
public node left;
public node right;
};
// Function to check if there exists a path
// with given product
// Strategy: divide the node value from the product
// if it is divisible when recurring down, and check
// to see if the product is 1 when you reach leaf
// node of the current path out of tree.
static bool hasPathProduct(node node, int prod)
{
// return true if we run out
// of tree and prod==1
if (node == null)
{
return (prod == 1);
}
else
{
bool ans = true;
// Check if product is divisible by
// current node, if not we are on wrong path
if (prod % (node.data) == 1)
{
return false;
}
// otherwise check both subtrees
int subProduct = prod / node.data;
// If we reach a leaf node and prod
// becomes 1 then return true
if (subProduct == 1 && node.left == null
&& node.right == null)
{
return true;
}
if (node.left != null)
{
ans = hasPathProduct(node.left, subProduct);
}
if (node.right != null)
{
ans = hasPathProduct(node.right, subProduct);
}
return ans;
}
}
/* UTILITY FUNCTIONS */
// Helper function that allocates
// a new node with the given data
// and null left and right pointers
static node newnode(int data)
{
node node = new node();
node.data = data;
node.left = null;
node.right = null;
return (node);
}
// Driver Code
public static void Main(String[] args)
{
int prod = 400;
/* Constructed binary tree is
10
/ \
8 2
/ \ /
3 5 2
*/
node root = newnode(10);
root.left = newnode(8);
root.right = newnode(2);
root.left.left = newnode(3);
root.left.right = newnode(5);
root.right.left = newnode(2);
if (hasPathProduct(root, prod))
{
Console.WriteLine("Yes");
}
else
{
Console.WriteLine("No");
}
}
}
// This code is contributed by Princi Singh
Javascript
<script>
// javascript program to check if there exists
// a root to leaf path product with
// given product // Binary Tree Node
class Node {
constructor(val) {
this.data = val;
this.left = null;
this.right = null;
}
}
// Function to check if there exists a path
// with given product
// Strategy: divide the node value from the product
// if it is divisible when recurring down, and check
// to see if the product is 1 when you reach leaf
// node of the current path out of tree.
function hasPathProduct( node , prod)
{
// return true if we run out
// of tree and prod==1
if (node == null)
{
return (prod == 1);
}
else
{
var ans = true;
// Check if product is divisible by
// current node, if not we are on wrong path
if (prod % (node.data) == 1)
{
return false;
}
// otherwise check both subtrees
var subProduct = prod / node.data;
// If we reach a leaf node and prod
// becomes 1 then return true
if (subProduct == 1 && node.left == null
&& node.right == null)
{
return true;
}
if (node.left != null)
{
ans = hasPathProduct(node.left, subProduct);
}
if (node.right != null)
{
ans = hasPathProduct(node.right, subProduct);
}
return ans;
}
}
/* UTILITY FUNCTIONS */
// Helper function that allocates
// a new node with the given data
// and null left and right pointers
function newnode(data)
{
var node = new Node();
node.data = data;
node.left = null;
node.right = null;
return (node);
}
// Driver Code
var prod = 400;
/* Constructed binary tree is
10
/ \
8 2
/ \ /
3 5 2
*/
var root = newnode(10);
root.left = newnode(8);
root.right = newnode(2);
root.left.left = newnode(3);
root.left.right = newnode(5);
root.right.left = newnode(2);
if (hasPathProduct(root, prod))
{
document.write("Yes");
}
else
{
document.write("No");
}
// This code contributed by Rajput-Ji
</script>
Salida :
YES
Tiempo Complejidad : O(n)
Espacio Auxiliar : O(n)
Publicación traducida automáticamente
Artículo escrito por VishalBachchas y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA