给定一个二叉树，任务是打印具有两个孩子的节点的数量，并且他们都是它的主要因素。
例子：
Input:
1
/ \
15 20
/ \ / \
3 5 4 2
\ /
2 3
Output: 1
Explanation:
Children of 15 (3, 5) are prime factors of 15
Input:
7
/ \
210 14
/ \ \
70 14 30
/ \ / \
2 5 3 5
/
23
Output: 2
Explanation:
Children of 70 (2, 5) are prime factors of 70
Children of 30 (3, 5) are prime factors of 30
方法：
- 遍历给定的二叉树，对于每个节点，检查两个孩子是否存在。
- 如果两个孩子都存在，检查每个孩子是否是这个节点的主要因素。
- 保留此类节点的计数并在最后打印。
- 为了检查一个因子是否为素数，我们将使用 Sieve 预先计算素数以在 O(1) 中进行检查。
下面是上述方法的实现。
C++
// C++ program for Counting nodes
// whose immediate children are its factors
#include
using namespace std;
int N = 1000000;
// To store all prime numbers
vector prime;
void SieveOfEratosthenes()
{
// Create a boolean array "prime[0..N]"
// and initialize all entries it as true.
// A value in prime[i] will finally
// be false if i is Not a prime, else true.
bool check[N + 1];
memset(check, true, sizeof(check));
for (int p = 2; p * p <= N; p++) {
// If prime[p] is not changed,
// then it is a prime
if (check[p] == true) {
prime.push_back(p);
// Update all multiples of p
// greater than or equal to
// the square of it
// numbers which are multiples of p
// and are less than p^2
// are already marked.
for (int i = p * p; i <= N; i += p)
check[i] = false;
}
}
}
// A Tree node
struct Node {
int key;
struct Node *left, *right;
};
// Utility function to create a new node
Node* newNode(int key)
{
Node* temp = new Node;
temp->key = key;
temp->left = temp->right = NULL;
return (temp);
}
// function to check and print if
// immediate children of a node
// are its factors or not
bool areChilrenPrimeFactors(struct Node* parent,
struct Node* a,
struct Node* b)
{
if (prime[a->key] && prime[b->key]
&& (parent->key % a->key == 0
&& parent->key % b->key == 0))
return true;
else
return false;
}
// Function to get the count of full Nodes in
// a binary tree
unsigned int getCount(struct Node* node)
{
// If tree is empty
if (!node)
return 0;
queue q;
// Initialize count of ful/l nodes
// having children as their factors
int count = 0;
// Do level order traversal
// starting from root
q.push(node);
while (!q.empty()) {
struct Node* temp = q.front();
q.pop();
if (temp->left && temp->right) {
if (areChilrenPrimeFactors(
temp, temp->left,
temp->right))
count++;
}
if (temp->left != NULL)
q.push(temp->left);
if (temp->right != NULL)
q.push(temp->right);
}
return count;
}
// Function to find total no of nodes
// In a given binary tree
int findSize(struct Node* node)
{
// Base condition
if (node == NULL)
return 0;
return 1
+ findSize(node->left)
+ findSize(node->right);
}
// Driver Code
int main()
{
/* 10
/ \
2 5
/ \
18 12
/ \ / \
2 3 3 2
/
7
*/
// Create Binary Tree as shown
Node* root = newNode(10);
root->left = newNode(2);
root->right = newNode(5);
root->right->left = newNode(18);
root->right->right = newNode(12);
root->right->left->left = newNode(2);
root->right->left->right = newNode(3);
root->right->right->left = newNode(3);
root->right->right->right = newNode(2);
root->right->right->right->left = newNode(7);
// To save all prime numbers
SieveOfEratosthenes();
// Print all nodes having
// children as their factors
cout << getCount(root) << endl;
return 0;
} Java
// Java program for Counting nodes
// whose immediate children are its factors
import java.util.*;
class GFG{
static int N = 1000000;
// To store all prime numbers
static Vector prime = new Vector();
static void SieveOfEratosthenes()
{
// Create a boolean array "prime[0..N]"
// and initialize all entries it as true.
// A value in prime[i] will finally
// be false if i is Not a prime, else true.
boolean []check = new boolean[N + 1];
Arrays.fill(check, true);
for (int p = 2; p * p <= N; p++) {
// If prime[p] is not changed,
// then it is a prime
if (check[p] == true) {
prime.add(p);
// Update all multiples of p
// greater than or equal to
// the square of it
// numbers which are multiples of p
// and are less than p^2
// are already marked.
for (int i = p * p; i <= N; i += p)
check[i] = false;
}
}
}
// A Tree node
static class Node {
int key;
Node left, right;
};
// Utility function to create a new node
static Node newNode(int key)
{
Node temp = new Node();
temp.key = key;
temp.left = temp.right = null;
return (temp);
}
// function to check and print if
// immediate children of a node
// are its factors or not
static boolean areChilrenPrimeFactors(Node parent,
Node a,
Node b)
{
if (prime.get(a.key) != null && prime.get(b.key) != null
&& (parent.key % a.key == 0
&& parent.key % b.key == 0))
return true;
else
return false;
}
// Function to get the count of full Nodes in
// a binary tree
static int getCount(Node node)
{
// If tree is empty
if (node == null)
return 0;
Queue q = new LinkedList<>();
// Initialize count of ful/l nodes
// having children as their factors
int count = 0;
// Do level order traversal
// starting from root
q.add(node);
while (!q.isEmpty()) {
Node temp = q.peek();
q.remove();
if (temp.left!=null && temp.right != null) {
if (areChilrenPrimeFactors(
temp, temp.left,
temp.right))
count++;
}
if (temp.left != null)
q.add(temp.left);
if (temp.right != null)
q.add(temp.right);
}
return count;
}
// Function to find total no of nodes
// In a given binary tree
static int findSize(Node node)
{
// Base condition
if (node == null)
return 0;
return 1
+ findSize(node.left)
+ findSize(node.right);
}
// Driver Code
public static void main(String[] args)
{
/* 10
/ \
2 5
/ \
18 12
/ \ / \
2 3 3 2
/
7
*/
// Create Binary Tree as shown
Node root = newNode(10);
root.left = newNode(2);
root.right = newNode(5);
root.right.left = newNode(18);
root.right.right = newNode(12);
root.right.left.left = newNode(2);
root.right.left.right = newNode(3);
root.right.right.left = newNode(3);
root.right.right.right = newNode(2);
root.right.right.right.left = newNode(7);
// To save all prime numbers
SieveOfEratosthenes();
// Print all nodes having
// children as their factors
System.out.print(getCount(root) +"\n");
}
}
// This code is contributed by Rajput-Ji C#
// C# program for Counting nodes
// whose immediate children are its factors
using System;
using System.Collections.Generic;
class GFG{
static int N = 1000000;
// To store all prime numbers
static List prime = new List();
static void SieveOfEratosthenes()
{
// Create a bool array "prime[0..N]"
// and initialize all entries it as true.
// A value in prime[i] will finally
// be false if i is Not a prime, else true.
bool []check = new bool[N + 1];
for (int i = 0; i <= N; i += 1){
check[i] = true;
}
for (int p = 2; p * p <= N; p++) {
// If prime[p] is not changed,
// then it is a prime
if (check[p] == true) {
prime.Add(p);
// Update all multiples of p
// greater than or equal to
// the square of it
// numbers which are multiples of p
// and are less than p^2
// are already marked.
for (int i = p * p; i <= N; i += p)
check[i] = false;
}
}
}
// A Tree node
class Node {
public int key;
public Node left, right;
};
// Utility function to create a new node
static Node newNode(int key)
{
Node temp = new Node();
temp.key = key;
temp.left = temp.right = null;
return (temp);
}
// function to check and print if
// immediate children of a node
// are its factors or not
static bool areChilrenPrimeFactors(Node parent,
Node a,
Node b)
{
if (prime.Contains(a.key)&& prime.Contains(b.key)
&& (parent.key % a.key == 0
&& parent.key % b.key == 0))
return true;
else
return false;
}
// Function to get the count of full Nodes in
// a binary tree
static int getCount(Node node)
{
// If tree is empty
if (node == null)
return 0;
List q = new List();
// Initialize count of ful/l nodes
// having children as their factors
int count = 0;
// Do level order traversal
// starting from root
q.Add(node);
while (q.Count!=0) {
Node temp = q[0];
q.RemoveAt(0);
if (temp.left!=null && temp.right != null) {
if (areChilrenPrimeFactors(
temp, temp.left,
temp.right))
count++;
}
if (temp.left != null)
q.Add(temp.left);
if (temp.right != null)
q.Add(temp.right);
}
return count;
}
// Function to find total no of nodes
// In a given binary tree
static int findSize(Node node)
{
// Base condition
if (node == null)
return 0;
return 1
+ findSize(node.left)
+ findSize(node.right);
}
// Driver Code
public static void Main(String[] args)
{
/* 10
/ \
2 5
/ \
18 12
/ \ / \
2 3 3 2
/
7
*/
// Create Binary Tree as shown
Node root = newNode(10);
root.left = newNode(2);
root.right = newNode(5);
root.right.left = newNode(18);
root.right.right = newNode(12);
root.right.left.left = newNode(2);
root.right.left.right = newNode(3);
root.right.right.left = newNode(3);
root.right.right.right = newNode(2);
root.right.right.right.left = newNode(7);
// To save all prime numbers
SieveOfEratosthenes();
// Print all nodes having
// children as their factors
Console.Write(getCount(root) +"\n");
}
}
// This code is contributed by Rajput-Ji 输出：
3
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