给定一棵二叉树，任务是打印其直接子节点为其因子的节点数。
例子：
Input:
1
/ \
15 20
/ \ / \
3 5 4 2
\ /
2 3
Output: 2
Explanation:
Children of 15 (3, 5)
are factors of 15
Children of 20 (4, 2)
are factors of 20
Input:
7
/ \
210 14
/ \ \
70 14 30
/ \ / \
2 5 10 15
/
23
Output:3
Explanation:
Children of 210 (70, 14)
are factors of 210
Children of 70 (2, 5)
are factors of 70
Children of 30 (10, 15)
are factors of 30
方法：为了解决这个问题，我们需要以 Level Order 方式遍历给定的二叉树，对于每个有两个孩子的节点，检查两个孩子是否都有作为当前节点值的因子的值。如果为真，则计算这些节点并在最后打印。
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;
// 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 areChilrenFactors(
struct Node* parent,
struct Node* a,
struct Node* b)
{
if (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;
// Do level order traversal
// starting from root
int count = 0;
// Store the number of nodes
// with both children as factors
q.push(node);
while (!q.empty()) {
struct Node* temp = q.front();
q.pop();
if (temp->left && temp->right) {
if (areChilrenFactors(
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
/ \
40 36
/ \
18 12
/ \ / \
2 6 3 4
/
7
*/
// Create Binary Tree as shown
Node* root = newNode(10);
root->left = newNode(40);
root->right = newNode(36);
root->right->left = newNode(18);
root->right->right = newNode(12);
root->right->left->left = newNode(2);
root->right->left->right = newNode(6);
root->right->right->left = newNode(3);
root->right->right->right = newNode(4);
root->right->right->right->left = newNode(7);
// 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{
// 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 areChilrenFactors(
Node parent,
Node a,
Node b)
{
if (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();
// Do level order traversal
// starting from root
int count = 0;
// Store the number of nodes
// with both children as factors
q.add(node);
while (!q.isEmpty()) {
Node temp = q.peek();
q.remove();
if (temp.left!=null && temp.right!=null) {
if (areChilrenFactors(
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
/ \
40 36
/ \
18 12
/ \ / \
2 6 3 4
/
7
*/
// Create Binary Tree as shown
Node root = newNode(10);
root.left = newNode(40);
root.right = newNode(36);
root.right.left = newNode(18);
root.right.right = newNode(12);
root.right.left.left = newNode(2);
root.right.left.right = newNode(6);
root.right.right.left = newNode(3);
root.right.right.right = newNode(4);
root.right.right.right.left = newNode(7);
// Print all nodes having
// children as their factors
System.out.print(getCount(root) +"\n");
}
}
// This code is contributed by sapnasingh4991 Python3
# Python3 program for counting nodes
# whose immediate children
# are its factors
from collections import deque as queue
# A Binary Tree Node
class Node:
def __init__(self, key):
self.data = key
self.left = None
self.right = None
# Function to check and print if
# immediate children of a node
# are its factors or not
def areChildrenFactors(parent, a, b):
if (parent.data % a.data == 0 and
parent.data % b.data == 0):
return True
else:
return False
# Function to get the
# count of full Nodes in
# a binary tree
def getCount(node):
# Base Case
if (not node):
return 0
q = queue()
# Do level order traversal
# starting from root
count = 0
# Store the number of nodes
# with both children as factors
q.append(node)
while (len(q) > 0):
temp = q.popleft()
#q.pop()
if (temp.left and temp.right):
if (areChildrenFactors(temp, temp.left,
temp.right)):
count += 1
if (temp.left != None):
q.append(temp.left)
if (temp.right != None):
q.append(temp.right)
return count
# Function to find total
# number of nodes
# In a given binary tree
def findSize(node):
# Base condition
if (node == None):
return 0
return (1 + findSize(node.left) +
findSize(node.right))
# Driver Code
if __name__ == '__main__':
# /* 10
# / \
# 40 36
# / \
# 18 12
# / \ / \
# 2 6 3 4
# /
# 7
# */
# Create Binary Tree
root = Node(10)
root.left = Node(40)
root.right = Node(36)
root.right.left = Node(18)
root.right.right = Node(12)
root.right.left.left = Node(2)
root.right.left.right = Node(6)
root.right.right.left = Node(3)
root.right.right.right = Node(4)
root.right.right.right.left = Node(7)
# Print all nodes having
# children as their factors
print(getCount(root))
# This code is contributed by mohit kumar 29C#
// C# program for Counting nodes
// whose immediate children
// are its factors
using System;
using System.Collections.Generic;
class GFG{
// 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 areChilrenFactors(
Node parent,
Node a,
Node b)
{
if (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();
// Do level order traversal
// starting from root
int count = 0;
// Store the number of nodes
// with both children as factors
q.Add(node);
while (q.Count != 0) {
Node temp = q[0];
q.RemoveAt(0);
if (temp.left!=null && temp.right != null) {
if (areChilrenFactors(
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
/ \
40 36
/ \
18 12
/ \ / \
2 6 3 4
/
7
*/
// Create Binary Tree as shown
Node root = newNode(10);
root.left = newNode(40);
root.right = newNode(36);
root.right.left = newNode(18);
root.right.right = newNode(12);
root.right.left.left = newNode(2);
root.right.left.right = newNode(6);
root.right.right.left = newNode(3);
root.right.right.right = newNode(4);
root.right.right.right.left = newNode(7);
// Print all nodes having
// children as their factors
Console.Write(getCount(root) +"\n");
}
}
// This code is contributed by sapnasingh4991 输出：
3
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