Root to leaf path sum equal to a given number

/*
    C Program 
    Root to leaf path sum equal to a given number
*/
#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 new node
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;
}
//Display pre order elements
void print_preorder(struct Node *node)
{
	if (node != NULL)
	{
		//Print node value
		printf("  %d", node->data);
		print_preorder(node->left);
		print_preorder(node->right);
	}
}
//Check whether the path of leaf nodes from the root is equal to the given sum?
int find_path_sum(struct Node *node, int sum, int k)
{
	if (node == NULL)
	{
		return 0;
	}
	if (node->left == NULL && node->right == NULL && k + node->data == sum)
	{
		// When current node is leaf node and path is equal to given sum
		return 1;
	}
	//Recursively executing left and right subtree
	if (find_path_sum(node->left, sum, node->data + k) || find_path_sum(node->right, sum, node->data + k))
	{
		return 1;
	}
	else
	{
		return 0;
	}
}
// Handles the request to find the sum exist in path from root to leaf nodes
void check_sum_path(struct Node *root, int sum)
{
	if (root == NULL)
	{
		printf("\n Empty Tree \n");
		return;
	}
	if (find_path_sum(root, sum, 0))
	{
		printf("\n Path from root to leaf of [%d] sum exists", sum);
	}
	else
	{
		printf("\n Path from root to leaf of [%d] sum are not exists", sum);
	}
}
int main()
{
	// Define pointer
	struct Node *root = NULL;
	/* 
	     3
	    /  \
	   /    \
	  2      1
	   \    /  \
	    1  5    6
	   / \
	  8   7
	-----------------------
	Binary Tree
	*/
	root = get_node(3);
	root->left = get_node(2);
	root->left->right = get_node(1);
	root->right = get_node(1);
	root->right->right = get_node(6);
	root->right->left = get_node(5);
	root->left->right->left = get_node(8);
	root->left->right->right = get_node(7);
	printf("\n Tree Node : \n");
	print_preorder(root);
	// Test Cases
	check_sum_path(root, 9);
	check_sum_path(root, 13);
	check_sum_path(root, 15);
	return 0;
}

Output

 Tree Node :
  3  2  1  8  7  1  5  6
 Path from root to leaf of [9] sum exists
 Path from root to leaf of [13] sum exists
 Path from root to leaf of [15] sum are not exists