# Count complete odd nodes path of binary tree in node js

Js program for Count complete odd nodes path of binary tree. Here mentioned other language solution.

``````/*
Node JS program for
Count odd paths in Binary Tree
*/
// Binary Tree node
class TreeNode
{
constructor(data)
{
// Set node value
this.data = data;
this.left = null;
this.right = null;
}
}
class BinaryTree
{
constructor()
{
this.root = null;
}
// Count all paths from root to leaf
// which containing all Odd nodes
countOddNodePath(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);
}
}
}

function main()
{
// 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.log(tree.countOddNodePath(tree.root));
}
// Start program execution
main();``````

Output

``3``

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