计算二叉树中的偶数路径

给定二叉树，任务是计算给定二叉树中偶数路径的数量。偶数路径是根到叶路径仅包含所有偶数节点的路径。
例子：

Input: Below is the given Binary Tree: 
 

Output: 3 
Explanation: 
There are 3 even path for the above Binary Tree: 
1. 10->12->2 
2. 10->4->18->22 
3. 10->4->18->24
Input: Below is the given Binary Tree: 
 

Output: 2 
Explanation: 
There are 2 even path for the above Binary Tree: 
1. 8->2->4 
2. 8->16->6->28

天真的方法：这个想法是生成从根到叶的所有路径，并检查每个路径中的所有节点是否均等。计算其中包含偶数节点的所有路径，然后返回计数。上面的实现占用了额外的空间来存储路径。
高效方法：想法是使用预排序树遍历。在对给定的二叉树进行遍历时，请执行以下操作：

1. 如果节点的当前值是奇数或指针变为NULL，则返回计数。
2. 如果当前节点是叶节点，则将计数增加1。
3. 递归调用与更新的计左和右子树。
4. 在所有递归调用之后，count的值是给定二叉树的偶数路径数。

下面是上述方法的实现：

C++

// C++ program for the above approach
#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);
}
 
// Utility function to count the even path
// in a given Binary tree
int evenPaths(struct Node* node, int count)
{
 
    // Base Condition, when node pointer
    // becomes null or node value is odd
    if (node == NULL || (node->key % 2 != 0)) {
        return count;
    }
 
    // Increment count when encounter leaf
    // node with all node value even
    if (!node->left && !node->right) {
        count++;
    }
 
    // Left recursive call, and save the
    // value of count
    count = evenPaths(node->left, count);
 
    // Right reursive call, and return
    // value of count
    return evenPaths(node->right, count);
}
 
// Function to count the even paths in a
// given Binary tree
int countEvenPaths(struct Node* node)
{
 
    // Function call with count = 0
    return evenPaths(node, 0);
}
 
// Driver Code
int main()
{
 
    // Tree
    Node* root = newNode(12);
    root->left = newNode(13);
    root->right = newNode(12);
 
    root->right->left = newNode(14);
    root->right->right = newNode(16);
 
    root->right->left->left = newNode(21);
    root->right->left->right = newNode(22);
    root->right->right->left = newNode(22);
    root->right->right->right = newNode(24);
    root->right->right->right->left = newNode(8);
 
    // Function call
    cout << countEvenPaths(root);
    return 0;
}

Java

// Java program for the above approach
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);
    }
     
    // Utility function to count the even path
    // in a given Binary tree
    static int evenPaths(Node node, int count)
    {
     
        // Base Condition, when node pointer
        // becomes null or node value is odd
        if (node == null || (node.key % 2 != 0)) {
            return count;
        }
     
        // Increment count when encounter leaf
        // node with all node value even
        if (node.left == null && node.right == null) {
            count++;
        }
     
        // Left recursive call, and save the
        // value of count
        count = evenPaths(node.left, count);
     
        // Right reursive call, and return
        // value of count
        return evenPaths(node.right, count);
    }
     
    // Function to count the even paths in a
    // given Binary tree
    static int countEvenPaths(Node node)
    {
     
        // Function call with count = 0
        return evenPaths(node, 0);
    }
     
    // Driver Code
    public static void main(String args[])
    {
     
        // Tree
        Node root = newNode(12);
        root.left = newNode(13);
        root.right = newNode(12);
     
        root.right.left = newNode(14);
        root.right.right = newNode(16);
     
        root.right.left.left = newNode(21);
        root.right.left.right = newNode(22);
        root.right.right.left = newNode(22);
        root.right.right.right = newNode(24);
        root.right.right.right.left = newNode(8);
     
        // Function call
        System.out.println(countEvenPaths(root));
         
    }
}
 
// This code is contributed by AbhiThakur

Python3

# Python3 program for the
# above approach
 
 
# A Tree node
class Node:
   
    def __init__(self, x):
       
        self.key = x
        self.left = None
        self.right = None
 
# Utility function to count
# the even path in a given
# Binary tree
def evenPaths(node, count):
   
    # Base Condition, when node
    # pointer becomes null or
    # node value is odd
    if (node == None or
       (node.key % 2 != 0)):
        return count
 
    # Increment count when
    # encounter leaf node
    # with all node value even
    if (not node.left and
        not node.right):
        count+=1
 
    # Left recursive call, and
    # save the value of count
    count = evenPaths(node.left,
                      count)
 
    # Right reursive call, and
    # return value of count
    return evenPaths(node.right,
                     count)
 
# Function to count the even
# paths in a given Binary tree
def countEvenPaths(node):
   
    # Function call with count = 0
    return evenPaths(node, 0)
 
# Driver Code
if __name__ == '__main__':
 
    #Tree
    root = Node(12)
    root.left = Node(13)
    root.right = Node(12)
 
    root.right.left = Node(14)
    root.right.right = Node(16)
 
    root.right.left.left = Node(21)
    root.right.left.right = Node(22)
    root.right.right.left = Node(22)
    root.right.right.right = Node(24)
    root.right.right.right.left = Node(8)
 
    #Function call
    print(countEvenPaths(root))
 
# This code is contributed by Mohit Kumar 29

C#

// C# program for the above approach
using System;
 
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);
    }
      
    // Utility function to count the even path
    // in a given Binary tree
    static int evenPaths(Node node, int count)
    {
      
        // Base Condition, when node pointer
        // becomes null or node value is odd
        if (node == null || (node.key % 2 != 0)) {
            return count;
        }
      
        // Increment count when encounter leaf
        // node with all node value even
        if (node.left == null && node.right == null) {
            count++;
        }
      
        // Left recursive call, and save the
        // value of count
        count = evenPaths(node.left, count);
      
        // Right reursive call, and return
        // value of count
        return evenPaths(node.right, count);
    }
      
    // Function to count the even paths in a
    // given Binary tree
    static int countEvenPaths(Node node)
    {
      
        // Function call with count = 0
        return evenPaths(node, 0);
    }
      
    // Driver Code
    public static void Main(String []args)
    {
      
        // Tree
        Node root = newNode(12);
        root.left = newNode(13);
        root.right = newNode(12);
      
        root.right.left = newNode(14);
        root.right.right = newNode(16);
      
        root.right.left.left = newNode(21);
        root.right.left.right = newNode(22);
        root.right.right.left = newNode(22);
        root.right.right.right = newNode(24);
        root.right.right.right.left = newNode(8);
      
        // Function call
        Console.WriteLine(countEvenPaths(root));
          
    }
}
 
// This code is contributed by PrinciRaj1992

输出：

3

时间复杂度： O(N)，其中N是给定二叉树中节点的数量。