给定一棵二叉树，任务是打印其直接子节点中的一个是其主要因子的节点数。
例子：
Input:
1
/ \
15 20
/ \ / \
3 5 4 2
\ /
2 3
Output: 3
Explanation:
Children of 15 (3, 5)
are prime factors of 15
Child of 20 (2)
is prime factors of 20
Child of 4 (2)
is prime factors of 4
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：
- 遍历给定的二叉树和每个节点：
- 检查孩子是否存在。
- 如果孩子存在，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，我们将使用埃拉托色尼筛法预先计算素数以在 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 its entries 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 if immediate children
// of a node are its factors or not
bool IsChilrenPrimeFactor(struct Node* parent,
struct Node* a)
{
if (prime[a->key]
&& (parent->key % a->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;
// Do level order traversal starting
// from root
// Initialize count of full nodes having
// children as their factors
int count = 0;
q.push(node);
while (!q.empty()) {
struct Node* temp = q.front();
q.pop();
// If only right child exist
if (temp->left == NULL
&& temp->right != NULL) {
if (IsChilrenPrimeFactor(
temp,
temp->right))
count++;
}
else {
// If only left child exist
if (temp->right == NULL
&& temp->left != NULL) {
if (IsChilrenPrimeFactor(
temp,
temp->left))
count++;
}
else {
// Both left and right child exist
if (temp->left != NULL
&& temp->right != NULL) {
if (IsChilrenPrimeFactor(
temp,
temp->right)
&& IsChilrenPrimeFactor(
temp,
temp->left))
count++;
}
}
}
// Check for left child
if (temp->left != NULL)
q.push(temp->left);
// Check for right child
if (temp->right != NULL)
q.push(temp->right);
}
return count;
}
// Driver Code
int main()
{
/* 10
/ \
2 5
/ \
18 12
/ \ / \
2 3 3 14
/
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(14);
root->right->right->right->left = newNode(7);
// To save all prime numbers
SieveOfEratosthenes();
// Print Count of 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 its entries 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 if immediate children
// of a node are its factors or not
static boolean IsChilrenPrimeFactor(Node parent,
Node a)
{
if (prime.get(a.key)>0
&& (parent.key % a.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<>();
// Do level order traversal starting
// from root
// Initialize count of full nodes having
// children as their factors
int count = 0;
q.add(node);
while (!q.isEmpty()) {
Node temp = q.peek();
q.remove();
// If only right child exist
if (temp.left == null
&& temp.right != null) {
if (IsChilrenPrimeFactor(
temp,
temp.right))
count++;
}
else {
// If only left child exist
if (temp.right == null
&& temp.left != null) {
if (IsChilrenPrimeFactor(
temp,
temp.left))
count++;
}
else {
// Both left and right child exist
if (temp.left != null
&& temp.right != null) {
if (IsChilrenPrimeFactor(
temp,
temp.right)
&& IsChilrenPrimeFactor(
temp,
temp.left))
count++;
}
}
}
// Check for left child
if (temp.left != null)
q.add(temp.left);
// Check for right child
if (temp.right != null)
q.add(temp.right);
}
return count;
}
// Driver Code
public static void main(String[] args)
{
/* 10
/ \
2 5
/ \
18 12
/ \ / \
2 3 3 14
/
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(14);
root.right.right.right.left = newNode(7);
// To save all prime numbers
SieveOfEratosthenes();
// Print Count of all nodes having children
// as their factors
System.out.print(GetCount(root) +"\n");
}
}
// This code is contributed by Rajput-Ji Python3
# Python program for Counting nodes whose
# immediate children are its factors
from collections import deque
N = 1000000
# To store all prime numbers
prime = []
def SieveOfEratosthenes() -> None:
# Create a boolean array "prime[0..N]"
# and initialize all its entries as true.
# A value in prime[i] will finally
# be false if i is Not a prime, else true.
check = [True for _ in range(N + 1)]
p = 2
while p * p <= N:
# If prime[p] is not changed,
# then it is a prime
if (check[p]):
prime.append(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 i in range(p * p, N + 1, p):
check[i] = False
p += 1
# A Tree node
class Node:
def __init__(self, key: int) -> None:
self.key = key
self.left = None
self.right = None
# Function to check if immediate children
# of a node are its factors or not
def IsChilrenPrimeFactor(parent: Node, a: Node) -> bool:
if (prime[a.key] and (parent.key % a.key == 0)):
return True
else:
return False
# Function to get the count of full Nodes in
# a binary tree
def GetCount(node: Node) -> int:
# If tree is empty
if (not node):
return 0
q = deque()
# Do level order traversal starting
# from root
# Initialize count of full nodes having
# children as their factors
count = 0
q.append(node)
while q:
temp = q.popleft()
# If only right child exist
if (temp.left == None and temp.right != None):
if (IsChilrenPrimeFactor(temp, temp.right)):
count += 1
else:
# If only left child exist
if (temp.right == None and temp.left != None):
if (IsChilrenPrimeFactor(temp, temp.left)):
count += 1
else:
# Both left and right child exist
if (temp.left != None and temp.right != None):
if (IsChilrenPrimeFactor(temp, temp.right)
and IsChilrenPrimeFactor(temp, temp.left)):
count += 1
# Check for left child
if (temp.left != None):
q.append(temp.left)
# Check for right child
if (temp.right != None):
q.append(temp.right)
return count
# Driver Code
if __name__ == "__main__":
''' 10
/ \
2 5
/ \
18 12
/ \ / \
2 3 3 14
/
7
'''
# Create Binary Tree as shown
root = Node(10)
root.left = Node(2)
root.right = Node(5)
root.right.left = Node(18)
root.right.right = Node(12)
root.right.left.left = Node(2)
root.right.left.right = Node(3)
root.right.right.left = Node(3)
root.right.right.right = Node(14)
root.right.right.right.left = Node(7)
# To save all prime numbers
SieveOfEratosthenes()
# Print Count of all nodes having children
# as their factors
print(GetCount(root))
# This code is contributed by sanjeev2552C#
// C# program for Counting nodes whose
// immediate children are its factors
using System;
using System.Collections.Generic;
public 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 its entries 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 if immediate children
// of a node are its factors or not
static bool IsChilrenPrimeFactor(Node parent,
Node a)
{
if (prime[a.key]>0
&& (parent.key % a.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();
// Do level order traversal starting
// from root
// Initialize count of full nodes having
// children as their factors
int count = 0;
q.Add(node);
while (q.Count!=0) {
Node temp = q[0];
q.RemoveAt(0);
// If only right child exist
if (temp.left == null
&& temp.right != null) {
if (IsChilrenPrimeFactor(
temp,
temp.right))
count++;
}
else {
// If only left child exist
if (temp.right == null
&& temp.left != null) {
if (IsChilrenPrimeFactor(
temp,
temp.left))
count++;
}
else {
// Both left and right child exist
if (temp.left != null
&& temp.right != null) {
if (IsChilrenPrimeFactor(
temp,
temp.right)
&& IsChilrenPrimeFactor(
temp,
temp.left))
count++;
}
}
}
// Check for left child
if (temp.left != null)
q.Add(temp.left);
// Check for right child
if (temp.right != null)
q.Add(temp.right);
}
return count;
}
// Driver Code
public static void Main(String[] args)
{
/* 10
/ \
2 5
/ \
18 12
/ \ / \
2 3 3 14
/
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(14);
root.right.right.right.left = newNode(7);
// To save all prime numbers
SieveOfEratosthenes();
// Print Count of all nodes having children
// as their factors
Console.Write(GetCount(root) +"\n");
}
}
// This code contributed by Rajput-Ji 输出：
3复杂度分析：
- 时间复杂度： O(N)。
在 dfs 中，树的每个节点都被处理一次，因此如果树中总共有 N 个节点，则由于 bfs 的复杂性是 O(N)。此外，为了处理每个节点，使用了 SieveOfEratosthenes()函数，它也具有 O(sqrt(N)) 的复杂度，但由于该函数仅执行一次，因此不会影响整体时间复杂度。因此，时间复杂度为 O(N)。 - 辅助空间： O(N)。
额外的空间用于素数数组，因此空间复杂度为 O(N)。
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