# Count complete odd nodes path of binary tree in swift

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

``````import Foundation
/*
Swift 4 program for
Count odd paths in Binary Tree
*/
// Binary Tree node
class TreeNode
{
var data: Int;
var left: TreeNode? ;
var right: TreeNode? ;
init(_ data: Int)
{
// Set node value
self.data = data;
self.left = nil;
self.right = nil;
}
}
class BinaryTree
{
var root: TreeNode? ;
init()
{
self.root = nil;
}
// Count all paths from root to leaf
// which containing all Odd nodes
func countOddNodePath(_ node: TreeNode? ) -> Int
{
if (node == nil || node!.data % 2 == 0)
{
// When tree node is null or
// its contain even value
return 0;
}
else
{
if (node!.left == nil && node!.right == nil)
{
// When get leaf node And
// path contain all Odd nodes
return 1;
}
return self.countOddNodePath(node!.left) +
self.countOddNodePath(node!.right);
}
}
static func main()
{
// New of binary tree
let tree: BinaryTree = BinaryTree();
/*
Construct Binary Tree
-----------------------
5
/  \
/    \
3      9
/ \    / \
7   6  1   3
/ \    / \
10  3  7   4
*/
tree.root = TreeNode(5);
tree.root!.left = TreeNode(3);
tree.root!.right = TreeNode(9);
tree.root!.right!.right = TreeNode(3);
tree.root!.left!.right = TreeNode(6);
tree.root!.right!.left = TreeNode(1);
tree.root!.left!.left = TreeNode(7);
tree.root!.left!.left!.left = TreeNode(10);
tree.root!.left!.left!.right = TreeNode(3);
tree.root!.right!.left!.right = TreeNode(4);
tree.root!.right!.left!.left = 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.
print(tree.countOddNodePath(tree.root));
}
}
BinaryTree.main();``````

Output

``3``

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