# Count complete odd nodes path of binary tree in c C program for Count complete odd nodes path of binary tree. Here more solutions.

``````// Include header file
#include <stdio.h>
#include <stdlib.h>
/*
C program for
Count odd paths in Binary Tree
*/
// Binary Tree node
typedef struct TreeNode
{
// Define useful field of TreeNode
int data;
struct TreeNode * left;
struct TreeNode * right;
}TreeNode;

TreeNode * getTreeNode(int data)
{
// Create dynamic memory of TreeNode
TreeNode * ref = (TreeNode * ) malloc(sizeof(TreeNode));
if (ref == NULL)
{
// Failed to create memory
return NULL;
}
// Set node value
ref->data = data;
ref->left = NULL;
ref->right = NULL;
return ref;
}
typedef struct BinaryTree
{
// Define useful field of BinaryTree
struct TreeNode * root;
}BinaryTree;

BinaryTree * getBinaryTree()
{
// Create dynamic memory of BinaryTree
BinaryTree * ref =
(BinaryTree * ) malloc(sizeof(BinaryTree));
if (ref == NULL)
{
// Failed to create memory
return NULL;
}
//Set initial tree root
ref->root = NULL;
return ref;
}
// Count all paths from root to leaf
// which containing all Odd nodes
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 countOddNodePath(node->left) +
countOddNodePath(node->right);
}
}
int main()
{
// New of binary tree
BinaryTree * tree = getBinaryTree();
/*
Construct Binary Tree
-----------------------
5
/  \
/    \
3      9
/ \    / \
7   6  1   3
/ \    / \
10  3  7   4
*/
tree->root = getTreeNode(5);
tree->root->left = getTreeNode(3);
tree->root->right = getTreeNode(9);
tree->root->right->right = getTreeNode(3);
tree->root->left->right = getTreeNode(6);
tree->root->right->left = getTreeNode(1);
tree->root->left->left = getTreeNode(7);
tree->root->left->left->left = getTreeNode(10);
tree->root->left->left->right = getTreeNode(3);
tree->root->right->left->right = getTreeNode(4);
tree->root->right->left->left = getTreeNode(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.
printf("%d\n",countOddNodePath(tree->root));
}``````

Output

``3``

## Comment

Please share your knowledge to improve code and content standard. Also submit your doubts, and test case. We improve by your feedback. We will try to resolve your query as soon as possible.