Print Ancestors of a given node in Binary Tree

#include
#include
#include

using namespace std;

/* A binary tree node has data, pointer to left child
and a pointer to right child */
struct node
{
int data;
struct node* left;
struct node* right;
};

/* If target is present in tree, then prints the ancestors
and returns true, otherwise returns false. */
bool printAncestors(struct node *root, int target)
{
/* base cases */
if ( root == NULL )
return false;

if ( root->data == target )
return true;

/* If target is present in either left or right,
then print this node */
if ( printAncestors(root->left, target) ||
printAncestors(root->right, target) )
{
cout<data<<" ";
return true;
}

/* Else return false */
return false;
}

/* Helper function that allocates a new node with the
given data and NULL left and right pointers. */
struct node* newnode(int data)
{
struct node* node = (struct node*)
malloc(sizeof(struct node));
node->data = data;
node->left = NULL;
node->right = NULL;

return(node);
}

/* Driver program to test above functions*/
int main()
{

/* Constructed the following binary tree
1
/ \
2 3
/ \
4 5
/
7
*/
struct node *root = newnode(1);
root->left = newnode(2);
root->right = newnode(3);
root->left->left = newnode(4);
root->left->right = newnode(5);
root->left->left->left = newnode(7);

printAncestors(root, 7);

getchar();
return 0;
}

Output:
4 2 1

Time Complexity: O(n) where n is the number of nodes in the given Binary Tree.