# Count complete odd nodes path of binary tree in c#

Csharp program for Count complete odd nodes path of binary tree. Here more solutions.

``````// Include namespace system
using System;
/*
Csharp program for
Count odd paths in Binary Tree
*/
// Binary Tree node
public class TreeNode
{
public int data;
public TreeNode left;
public TreeNode right;
public TreeNode(int data)
{
// Set node value
this.data = data;
this.left = null;
this.right = null;
}
}
public class BinaryTree
{
public TreeNode root;
public BinaryTree()
{
//Set initial tree root
this.root = null;
}
// Count all paths from root to leaf
// which containing all Odd nodes
public int countOddNodePath(TreeNode node)
{
if (node == null || node.data % 2 == 0)
{
// When tree node is null or
// its contain even value
return 0;
}
else
{
if (node.left == null && node.right == null)
{
// When get leaf node And
// path contain all Odd nodes
return 1;
}
return this.countOddNodePath(node.left) +
this.countOddNodePath(node.right);
}
}
public static void Main(String[] args)
{
// New of binary tree
var tree = new BinaryTree();
/*
Construct Binary Tree
-----------------------
5
/  \
/    \
3      9
/ \    / \
7   6  1   3
/ \    / \
10  3  7   4
*/
tree.root = new TreeNode(5);
tree.root.left = new TreeNode(3);
tree.root.right = new TreeNode(9);
tree.root.right.right = new TreeNode(3);
tree.root.left.right = new TreeNode(6);
tree.root.right.left = new TreeNode(1);
tree.root.left.left = new TreeNode(7);
tree.root.left.left.left = new TreeNode(10);
tree.root.left.left.right = new TreeNode(3);
tree.root.right.left.right = new TreeNode(4);
tree.root.right.left.left = new TreeNode(7);
/*
Given Binary Tree
-----------------------
5
/  \
/    \
3      9
/ \    / \
7   6  1   3
/ \    / \
10  3  7   4
Here
Odd path from root to leaf
-----------------------
5
/  \
/    \
3      9
/      / \
7      1   3
\    /
3  7
---------------------
5->3->7->3
5->9->1->7
5->9->3
----------------------
Result : 3 [number of path]
*/
// Count odd nodes paths from root
// to leaf in a binary tree.
Console.WriteLine(tree.countOddNodePath(tree.root));
}
}``````

Output

``3``

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