具有奇数个偶数的子树数
给定一棵二叉树，求奇数个偶数个子树的个数。
例子：
Input : 2
/ \
1 3
/ \ / \
4 10 8 5
/
6
Output : 6
The subtrees are 4, 6, 1, 8, 3
/ \ / \
4 10 8 5
/
6
2
/ \
1 3
/ \ / \
4 10 8 5
/
6 这个想法是递归遍历树。对于每个节点，递归计算左右子树中的偶数。如果 root 也是偶数，则将其也添加到计数中。如果计数变为奇数，则递增结果。
count = 0 // Initialize result
int countSubtrees(root)
{
if (root == NULL)
return 0;
// count even numbers in left subtree
int c = countSubtrees(root-> left);
// Add count of even numbers in right subtree
c += countSubtrees(root-> right);
// check if root data is an even number
if (root-> data % 2 == 0)
c += 1;
// If total count of even numbers
// for the subtree is odd
if (c % 2 != 0)
count++;
// return total count of even
// numbers of the subtree
return c;
}C++
// C++ implementation to find number of
// subtrees having odd count of even numbers
#include
using namespace std;
/* A binary tree Node */
struct Node
{
int data;
struct Node* left, *right;
};
/* Helper function that allocates a new Node with the
given data and NULL left and right pointers. */
struct Node* newNode(int data)
{
struct Node* node = new Node;
node->data = data;
node->left = node->right = NULL;
return(node);
}
// Returns count of subtrees having odd count of
// even numbers
int countRec(struct Node* root, int *pcount)
{
// base condition
if (root == NULL)
return 0;
// count even nodes in left subtree
int c = countRec(root->left, pcount);
// Add even nodes in right subtree
c += countRec(root->right, pcount);
// Check if root data is an even number
if (root->data % 2 == 0)
c += 1;
// if total count of even numbers
// for the subtree is odd
if (c % 2 != 0)
(*pcount)++;
// Total count of even nodes of the subtree
return c;
}
// A wrapper over countRec()
int countSubtrees(Node *root)
{
int count = 0;
int *pcount = &count;
countRec(root, pcount);
return count;
}
// Driver program to test above
int main()
{
// binary tree formation
struct Node *root = newNode(2); /* 2 */
root->left = newNode(1); /* / \ */
root->right = newNode(3); /* 1 3 */
root->left->left = newNode(4); /* / \ / \ */
root->left->right = newNode(10); /* 4 10 8 5 */
root->right->left = newNode(8); /* / */
root->right->right = newNode(5); /* 6 */
root->left->right->left = newNode(6);
cout<<"Count = "<< countSubtrees(root);
return 0;
}
// This code is contributed by famously. C
// C implementation to find number of
// subtrees having odd count of even numbers
#include
/* A binary tree Node */
struct Node
{
int data;
struct Node* left, *right;
};
/* Helper function that allocates a new Node with the
given data and NULL left and right pointers. */
struct Node* newNode(int data)
{
struct Node* node = new Node;
node->data = data;
node->left = node->right = NULL;
return(node);
}
// Returns count of subtrees having odd count of
// even numbers
int countRec(struct Node* root, int *pcount)
{
// base condition
if (root == NULL)
return 0;
// count even nodes in left subtree
int c = countRec(root->left, pcount);
// Add even nodes in right subtree
c += countRec(root->right, pcount);
// Check if root data is an even number
if (root->data % 2 == 0)
c += 1;
// if total count of even numbers
// for the subtree is odd
if (c % 2 != 0)
(*pcount)++;
// Total count of even nodes of the subtree
return c;
}
// A wrapper over countRec()
int countSubtrees(Node *root)
{
int count = 0;
int *pcount = &count;
countRec(root, pcount);
return count;
}
// Driver program to test above
int main()
{
// binary tree formation
struct Node *root = newNode(2); /* 2 */
root->left = newNode(1); /* / \ */
root->right = newNode(3); /* 1 3 */
root->left->left = newNode(4); /* / \ / \ */
root->left->right = newNode(10); /* 4 10 8 5 */
root->right->left = newNode(8); /* / */
root->right->right = newNode(5); /* 6 */
root->left->right->left = newNode(6);
printf("Count = %d", countSubtrees(root));
return 0;
} Java
// Java implementation to find number of
// subtrees having odd count of even numbers
import java.util.*;
class GfG {
/* A binary tree Node */
static class Node
{
int data;
Node left, right;
}
/* 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);
}
// Returns count of subtrees having odd count of
// even numbers
static class P
{
int pcount = 0;
}
static int countRec(Node root, P p)
{
// base condition
if (root == null)
return 0;
// count even nodes in left subtree
int c = countRec(root.left, p);
// Add even nodes in right subtree
c += countRec(root.right, p);
// Check if root data is an even number
if (root.data % 2 == 0)
c += 1;
// if total count of even numbers
// for the subtree is odd
if (c % 2 != 0)
(p.pcount)++;
// Total count of even nodes of the subtree
return c;
}
// A wrapper over countRec()
static int countSubtrees(Node root)
{
P p = new P();
countRec(root, p);
return p.pcount;
}
// Driver program to test above
public static void main(String[] args)
{
// binary tree formation
Node root = newNode(2); /* 2 */
root.left = newNode(1); /* / \ */
root.right = newNode(3); /* 1 3 */
root.left.left = newNode(4); /* / \ / \ */
root.left.right = newNode(10); /* 4 10 8 5 */
root.right.left = newNode(8); /* / */
root.right.right = newNode(5); /* 6 */
root.left.right.left = newNode(6);
System.out.println("Count = " + countSubtrees(root));
}
}Python3
# Python3 implementation to find number of
# subtrees having odd count of even numbers
# Helper class 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
# Returns count of subtrees having odd
# count of even numbers
def countRec(root, pcount):
# base condition
if (root == None):
return 0
# count even nodes in left subtree
c = countRec(root.left, pcount)
# Add even nodes in right subtree
c += countRec(root.right, pcount)
# Check if root data is an even number
if (root.data % 2 == 0):
c += 1
# if total count of even numbers
# for the subtree is odd
if c % 2 != 0:
pcount[0] += 1
# Total count of even nodes of
# the subtree
return c
# A wrapper over countRec()
def countSubtrees(root):
count = [0]
pcount = count
countRec(root, pcount)
return count[0]
# Driver Code
if __name__ == '__main__':
# binary tree formation
root = newNode(2) # 2
root.left = newNode(1) # / \
root.right = newNode(3) # 1 3
root.left.left = newNode(4) # / \ / \
root.left.right = newNode(10) # 4 10 8 5
root.right.left = newNode(8) # /
root.right.right = newNode(5) # 6
root.left.right.left = newNode(6)
print("Count = ", countSubtrees(root))
# This code is contributed by PranchalKC#
// C# implementation to find number of
// subtrees having odd count of even numbers
using System;
class GFG
{
/* A binary tree Node */
public class Node
{
public int data;
public Node left, right;
}
/* 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);
}
// Returns count of subtrees
// having odd count of even numbers
public class P
{
public int pcount = 0;
}
static int countRec(Node root, P p)
{
// base condition
if (root == null)
return 0;
// count even nodes in left subtree
int c = countRec(root.left, p);
// Add even nodes in right subtree
c += countRec(root.right, p);
// Check if root data is an even number
if (root.data % 2 == 0)
c += 1;
// if total count of even numbers
// for the subtree is odd
if (c % 2 != 0)
(p.pcount)++;
// Total count of even nodes of the subtree
return c;
}
// A wrapper over countRec()
static int countSubtrees(Node root)
{
P p = new P();
countRec(root, p);
return p.pcount;
}
// Driver Code
public static void Main(String[] args)
{
// binary tree formation
Node root = newNode(2); /* 2 */
root.left = newNode(1); /* / \ */
root.right = newNode(3); /* 1 3 */
root.left.left = newNode(4); /* / \ / \ */
root.left.right = newNode(10); /* 4 10 8 5 */
root.right.left = newNode(8); /* / */
root.right.right = newNode(5); /* 6 */
root.left.right.left = newNode(6);
Console.WriteLine("Count = " +
countSubtrees(root));
}
}
// This code is contributed by 29AjayKumarJavascript
输出：
Count = 6