# Count Even paths in Binary Tree

Here given code implementation process.

```
// C program
// Count even paths in Binary Tree
#include <stdio.h>
#include <stdlib.h>
//Binary Tree node
struct Node
{
int data;
struct Node *left, *right;
};
//This is creating a binary tree node and return this
struct Node *get_node(int data)
{
// Create dynamic node
struct Node *new_node = (struct Node *) malloc(sizeof(struct Node));
if (new_node != NULL)
{
//Set data and pointer values
new_node->data = data;
new_node->left = NULL;
new_node->right = NULL;
}
else
{
//This is indicates, segmentation fault or memory overflow problem
printf("Memory Overflow\n");
}
//return new node
return new_node;
}
// Count all paths from root to leaf which containing all Even nodes
int count_even_path(struct Node *root)
{
if (root == NULL || root->data % 2 != 0)
{
//When tree node is null or its contain Odd value
return 0;
}
else
{
if (root->left == NULL && root->right == NULL)
{
//When get leaf node
return 1;
}
return count_even_path(root->left) + count_even_path(root->right);
}
}
int main()
{
struct Node *root = NULL;
/*
Construct Binary Tree
-----------------------
2
/ \
4 10
/ \ / \
7 6 12 3
/ \ / \ \
10 3 7 4 8
*/
//Add tree node
root = get_node(2);
root->left = get_node(4);
root->right = get_node(10);
root->right->right = get_node(3);
root->left->right = get_node(6);
root->right->left = get_node(12);
root->left->left = get_node(7);
root->left->left->left = get_node(10);
root->left->left->right = get_node(3);
root->right->left->right = get_node(4);
root->right->left->left = get_node(7);
root->right->right->right = get_node(8);
int counter = count_even_path(root);
// Display calculated result
printf("Total Even path in this tree are : %d \n", counter);
return 0;
}
```

#### Output

`Total Even path in this tree are : 2`

```
/*
Java Program
Count even paths in Binary Tree
*/
//Binary Tree node
class Node
{
public int data;
public Node left;
public Node right;
public Node(int data)
{
//set node value
this.data = data;
this.left = null;
this.right = null;
}
}
class BinaryTree
{
public Node root;
public BinaryTree()
{
//Set initial tree root to null
this.root = null;
}
// Count all paths from root to leaf which containing all Even nodes
public int count_even_path(Node root)
{
if (root == null || root.data % 2 != 0)
{
//When tree node is null or its contain Odd value
return 0;
}
else
{
if (root.left == null && root.right == null)
{
//When get leaf node
return 1;
}
return count_even_path(root.left) + count_even_path(root.right);
}
}
public static void main(String[] args)
{
//Make object of binary tree
BinaryTree tree = new BinaryTree();
/*
Construct Binary Tree
-----------------------
2
/ \
4 10
/ \ / \
7 6 12 3
/ \ / \ \
10 3 7 4 8
*/
//Add tree node
tree.root = new Node(2);
tree.root.left = new Node(4);
tree.root.right = new Node(10);
tree.root.right.right = new Node(3);
tree.root.left.right = new Node(6);
tree.root.right.left = new Node(12);
tree.root.left.left = new Node(7);
tree.root.left.left.left = new Node(10);
tree.root.left.left.right = new Node(3);
tree.root.right.left.right = new Node(4);
tree.root.right.left.left = new Node(7);
tree.root.right.right.right = new Node(8);
int counter = tree.count_even_path(tree.root);
// Display calculated result
System.out.print("Total Even path in this tree are : " + counter + " \n");
}
}
```

#### Output

`Total Even path in this tree are : 2`

```
//Include header file
#include <iostream>
using namespace std;
/*
C++ Program
Count even paths in Binary Tree
*/
//Binary Tree node
class Node
{
public: int data;
Node *left;
Node *right;
Node(int data)
{
//set node value
this->data = data;
this->left = NULL;
this->right = NULL;
}
};
class BinaryTree
{
public:
Node *root;
BinaryTree()
{
//Set initial tree root to null
this->root = NULL;
}
// Count all paths from root to leaf which containing all Even nodes
int count_even_path(Node *root)
{
if (root == NULL || root->data % 2 != 0)
{
//When tree node is null or its contain Odd value
return 0;
}
else
{
if (root->left == NULL && root->right == NULL)
{
//When get leaf node
return 1;
}
return this->count_even_path(root->left) + this->count_even_path(root->right);
}
}
};
int main()
{
//Make object of binary tree
BinaryTree tree = BinaryTree();
/*
Construct Binary Tree
-----------------------
2
/ \
4 10
/ \ / \
7 6 12 3
/ \ / \ \
10 3 7 4 8
*/
//Add tree node
tree.root = new Node(2);
tree.root->left = new Node(4);
tree.root->right = new Node(10);
tree.root->right->right = new Node(3);
tree.root->left->right = new Node(6);
tree.root->right->left = new Node(12);
tree.root->left->left = new Node(7);
tree.root->left->left->left = new Node(10);
tree.root->left->left->right = new Node(3);
tree.root->right->left->right = new Node(4);
tree.root->right->left->left = new Node(7);
tree.root->right->right->right = new Node(8);
int counter = tree.count_even_path(tree.root);
// Display calculated result
cout << "Total Even path in this tree are : " << counter << " \n";
return 0;
}
```

#### Output

`Total Even path in this tree are : 2`

```
//Include namespace system
using System;
/*
C# Program
Count even paths in Binary Tree
*/
//Binary Tree node
class Node
{
public int data;
public Node left;
public Node right;
public Node(int data)
{
//set node value
this.data = data;
this.left = null;
this.right = null;
}
}
class BinaryTree
{
public Node root;
public BinaryTree()
{
//Set initial tree root to null
this.root = null;
}
// Count all paths from root to leaf which containing all Even nodes
public int count_even_path(Node root)
{
if (root == null || root.data % 2 != 0)
{
//When tree node is null or its contain Odd value
return 0;
}
else
{
if (root.left == null && root.right == null)
{
//When get leaf node
return 1;
}
return count_even_path(root.left) + count_even_path(root.right);
}
}
public static void Main(String[] args)
{
//Make object of binary tree
BinaryTree tree = new BinaryTree();
/*
Construct Binary Tree
-----------------------
2
/ \
4 10
/ \ / \
7 6 12 3
/ \ / \ \
10 3 7 4 8
*/
//Add tree node
tree.root = new Node(2);
tree.root.left = new Node(4);
tree.root.right = new Node(10);
tree.root.right.right = new Node(3);
tree.root.left.right = new Node(6);
tree.root.right.left = new Node(12);
tree.root.left.left = new Node(7);
tree.root.left.left.left = new Node(10);
tree.root.left.left.right = new Node(3);
tree.root.right.left.right = new Node(4);
tree.root.right.left.left = new Node(7);
tree.root.right.right.right = new Node(8);
int counter = tree.count_even_path(tree.root);
// Display calculated result
Console.Write("Total Even path in this tree are : " + counter + " \n");
}
}
```

#### Output

`Total Even path in this tree are : 2`

```
<?php
/*
Php Program
Count even paths in Binary Tree
*/
//Binary Tree node
class Node
{
public $data;
public $left;
public $right;
function __construct($data)
{
//set node value
$this->data = $data;
$this->left = null;
$this->right = null;
}
}
class BinaryTree
{
public $root;
function __construct()
{
//Set initial tree root to null
$this->root = null;
}
// Count all paths from root to leaf which containing all Even nodes
public function count_even_path($root)
{
if ($root == null || $root->data % 2 != 0)
{
//When tree node is null or its contain Odd value
return 0;
}
else
{
if ($root->left == null && $root->right == null)
{
//When get leaf node
return 1;
}
return $this->count_even_path($root->left) + $this->count_even_path($root->right);
}
}
}
function main()
{
//Make object of binary tree
$tree = new BinaryTree();
/*
Construct Binary Tree
-----------------------
2
/ \
4 10
/ \ / \
7 6 12 3
/ \ / \ \
10 3 7 4 8
*/
//Add tree node
$tree->root = new Node(2);
$tree->root->left = new Node(4);
$tree->root->right = new Node(10);
$tree->root->right->right = new Node(3);
$tree->root->left->right = new Node(6);
$tree->root->right->left = new Node(12);
$tree->root->left->left = new Node(7);
$tree->root->left->left->left = new Node(10);
$tree->root->left->left->right = new Node(3);
$tree->root->right->left->right = new Node(4);
$tree->root->right->left->left = new Node(7);
$tree->root->right->right->right = new Node(8);
$counter = $tree->count_even_path($tree->root);
// Display calculated result
echo "Total Even path in this tree are : ". $counter ." \n";
}
main();
```

#### Output

`Total Even path in this tree are : 2`

```
/*
Node Js Program
Count even paths in Binary Tree
*/
//Binary Tree node
class Node
{
constructor(data)
{
//set node value
this.data = data;
this.left = null;
this.right = null;
}
}
class BinaryTree
{
constructor()
{
//Set initial tree root to null
this.root = null;
}
// Count all paths from root to leaf which containing all Even nodes
count_even_path(root)
{
if (root == null || root.data % 2 != 0)
{
//When tree node is null or its contain Odd value
return 0;
}
else
{
if (root.left == null && root.right == null)
{
//When get leaf node
return 1;
}
return this.count_even_path(root.left) + this.count_even_path(root.right);
}
}
}
function main()
{
//Make object of binary tree
var tree = new BinaryTree();
/*
Construct Binary Tree
-----------------------
2
/ \
4 10
/ \ / \
7 6 12 3
/ \ / \ \
10 3 7 4 8
*/
//Add tree node
tree.root = new Node(2);
tree.root.left = new Node(4);
tree.root.right = new Node(10);
tree.root.right.right = new Node(3);
tree.root.left.right = new Node(6);
tree.root.right.left = new Node(12);
tree.root.left.left = new Node(7);
tree.root.left.left.left = new Node(10);
tree.root.left.left.right = new Node(3);
tree.root.right.left.right = new Node(4);
tree.root.right.left.left = new Node(7);
tree.root.right.right.right = new Node(8);
var counter = tree.count_even_path(tree.root);
// Display calculated result
process.stdout.write("Total Even path in this tree are : " + counter + " \n");
}
main();
```

#### Output

`Total Even path in this tree are : 2`

```
# Python 3 Program
# Count even paths in Binary Tree
# Binary Tree node
class Node :
def __init__(self, data) :
# Set node value
self.data = data
self.left = None
self.right = None
class BinaryTree :
def __init__(self) :
# Set initial tree root to null
self.root = None
# Count all paths from root to leaf which containing all Even nodes
def count_even_path(self, root) :
if (root == None or root.data % 2 != 0) :
# When tree node is null or its contain Odd value
return 0
else :
if (root.left == None and root.right == None) :
# When get leaf node
return 1
return self.count_even_path(root.left) + self.count_even_path(root.right)
def main() :
# Make object of binary tree
tree = BinaryTree()
#
# Construct Binary Tree
# -----------------------
# 2
# / \
# 4 10
# / \ / \
# 7 6 12 3
# / \ / \ \
# 10 3 7 4 8
#
# Add tree node
tree.root = Node(2)
tree.root.left = Node(4)
tree.root.right = Node(10)
tree.root.right.right = Node(3)
tree.root.left.right = Node(6)
tree.root.right.left = Node(12)
tree.root.left.left = Node(7)
tree.root.left.left.left = Node(10)
tree.root.left.left.right = Node(3)
tree.root.right.left.right = Node(4)
tree.root.right.left.left = Node(7)
tree.root.right.right.right = Node(8)
counter = tree.count_even_path(tree.root)
# Display calculated result
print("Total Even path in this tree are : ", counter ," \n", end = "")
if __name__ == "__main__": main()
```

#### Output

`Total Even path in this tree are : 2`

```
# Ruby Program
# Count even paths in Binary Tree
# Binary Tree node
class Node
# Define the accessor and reader of class Node
attr_reader :data, :left, :right
attr_accessor :data, :left, :right
def initialize(data)
# Set node value
self.data = data
self.left = nil
self.right = nil
end
end
class BinaryTree
# Define the accessor and reader of class BinaryTree
attr_reader :root
attr_accessor :root
def initialize()
# Set initial tree root to null
self.root = nil
end
# Count all paths from root to leaf which containing all Even nodes
def count_even_path(root)
if (root == nil || root.data % 2 != 0)
# When tree node is null or its contain Odd value
return 0
else
if (root.left == nil && root.right == nil)
# When get leaf node
return 1
end
return self.count_even_path(root.left) + self.count_even_path(root.right)
end
end
end
def main()
# Make object of binary tree
tree = BinaryTree.new()
#
# Construct Binary Tree
# -----------------------
# 2
# / \
# 4 10
# / \ / \
# 7 6 12 3
# / \ / \ \
# 10 3 7 4 8
#
# Add tree node
tree.root = Node.new(2)
tree.root.left = Node.new(4)
tree.root.right = Node.new(10)
tree.root.right.right = Node.new(3)
tree.root.left.right = Node.new(6)
tree.root.right.left = Node.new(12)
tree.root.left.left = Node.new(7)
tree.root.left.left.left = Node.new(10)
tree.root.left.left.right = Node.new(3)
tree.root.right.left.right = Node.new(4)
tree.root.right.left.left = Node.new(7)
tree.root.right.right.right = Node.new(8)
counter = tree.count_even_path(tree.root)
# Display calculated result
print("Total Even path in this tree are : ", counter ," \n")
end
main()
```

#### Output

```
Total Even path in this tree are : 2
```

```
/*
Scala Program
Count even paths in Binary Tree
*/
//Binary Tree node
class Node(var data: Int,
var left: Node,
var right: Node)
{
def this(data: Int)
{
this(data, null, null);
}
}
class BinaryTree(var root: Node)
{
def this()
{
this(null);
}
// Count all paths from root to leaf which containing all Even nodes
def count_even_path(root: Node): Int = {
if (root == null || root.data % 2 != 0)
{
//When tree node is null or its contain Odd value
return 0;
}
else
{
if (root.left == null && root.right == null)
{
//When get leaf node
return 1;
}
return count_even_path(root.left) + count_even_path(root.right);
}
}
}
object Main
{
def main(args: Array[String]): Unit = {
//Make object of binary tree
var tree: BinaryTree = new BinaryTree();
/*
Construct Binary Tree
-----------------------
2
/ \
4 10
/ \ / \
7 6 12 3
/ \ / \ \
10 3 7 4 8
*/
//Add tree node
tree.root = new Node(2);
tree.root.left = new Node(4);
tree.root.right = new Node(10);
tree.root.right.right = new Node(3);
tree.root.left.right = new Node(6);
tree.root.right.left = new Node(12);
tree.root.left.left = new Node(7);
tree.root.left.left.left = new Node(10);
tree.root.left.left.right = new Node(3);
tree.root.right.left.right = new Node(4);
tree.root.right.left.left = new Node(7);
tree.root.right.right.right = new Node(8);
var counter: Int = tree.count_even_path(tree.root);
// Display calculated result
print("Total Even path in this tree are : " + counter + " \n");
}
}
```

#### Output

`Total Even path in this tree are : 2`

```
/*
Swift 4 Program
Count even paths in Binary Tree
*/
//Binary Tree node
class Node
{
var data: Int;
var left: Node? ;
var right: Node? ;
init(_ data: Int)
{
//Set node value
self.data = data;
self.left = nil;
self.right = nil;
}
}
class BinaryTree
{
var root: Node? ;
init()
{
//Set initial tree root to null
self.root = nil;
}
// Count all paths from root to leaf which containing all Even nodes
func count_even_path(_ root: Node? ) -> Int
{
if (root == nil || root!.data % 2 != 0)
{
//When tree node is null or its contain Odd value
return 0;
}
else
{
if (root!.left == nil && root!.right == nil)
{
//When get leaf node
return 1;
}
return self.count_even_path(root!.left) + self.count_even_path(root!.right);
}
}
}
func main()
{
//Make object of binary tree
let tree: BinaryTree = BinaryTree();
/*
Construct Binary Tree
-----------------------
2
/ \
4 10
/ \ / \
7 6 12 3
/ \ / \ \
10 3 7 4 8
*/
//Add tree node
tree.root = Node(2);
tree.root!.left = Node(4);
tree.root!.right = Node(10);
tree.root!.right!.right = Node(3);
tree.root!.left!.right = Node(6);
tree.root!.right!.left = Node(12);
tree.root!.left!.left = Node(7);
tree.root!.left!.left!.left = Node(10);
tree.root!.left!.left!.right = Node(3);
tree.root!.right!.left!.right = Node(4);
tree.root!.right!.left!.left = Node(7);
tree.root!.right!.right!.right = Node(8);
let counter: Int = tree.count_even_path(tree.root);
// Display calculated result
print("Total Even path in this tree are : ", counter ," \n", terminator: "");
}
main();
```

#### Output

`Total Even path in this tree are : 2`

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