Given a tree and a sum, return true if there is a path from the root down to a leaf, such that adding up all the values along the path equals the

#include
#include
#define bool int

/* 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;
};

/*
Given a tree and a sum, return true if there is a path from the root
down to a leaf, such that adding up all the values along the path
equals the given sum.

Strategy: subtract the node value from the sum when recurring down,
and check to see if the sum is 0 when you run out of tree.
*/
bool hasPathSum(struct node* node, int sum)
{
/* return true if we run out of tree and sum==0 */
if (node == NULL)
{
return(sum == 0);
}
else
{
/* otherwise check both subtrees */
int subSum = sum - node->data;
return(hasPathSum(node->left, subSum) ||
hasPathSum(node->right, subSum));
}
}

/* UTILITY FUNCTIONS */
/* 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()
{

int sum = 21;

/* Constructed binary tree is
10
/ \
8 2
/ \ /
3 5 2
*/
struct node *root = newnode(10);
root->left = newnode(8);
root->right = newnode(2);
root->left->left = newnode(3);
root->left->right = newnode(5);
root->right->left = newnode(2);

if(hasPathSum(root, sum))
printf("There is a root-to-leaf path with sum %d", sum);
else
printf("There is no root-to-leaf path with sum %d", sum);

getchar();
return 0;
}