On single parent use case (single char), just store the no. of chars

This saves more than huffman encoding. Note the header is still stored as usual.

tree.c

@@ -194,6 +194,13 @@ int encode_tree(struct BIT_BUFFER *bitbuf, struct tree *parent, char *buf,
 {
 {
 	printf("Bit pos at %d %d\n", ftell(bitbuf->fp), bitbuf->pos);
 	printf("Bit pos at %d %d\n", ftell(bitbuf->fp), bitbuf->pos);
 
 
+	/* It makes much more sense to write the count when there is only one
+	 * char. */
+	if (!parent->left && !parent->right) {
+		bb_writebyte(bitbuf, parent->freq);
+		return 0;
+	}
+
 	int maxheight = get_max_height(parent) - 1;
 	int maxheight = get_max_height(parent) - 1;
 	int *pathbuf = calloc(maxheight, sizeof(int));
 	int *pathbuf = calloc(maxheight, sizeof(int));
 
 
@@ -220,16 +227,14 @@ static int read_entries(struct BIT_BUFFER *bitbuf, FILE *out,
 
 
 	/* If we compress only one letter, the root node is the node containing
 	/* If we compress only one letter, the root node is the node containing
 	 * character. So this needs special care. */
 	 * character. So this needs special care. */
-	#if 0
 	if (!parent->left && !parent->right) {
 	if (!parent->left && !parent->right) {
-		while (ftell(bitbuf->fp) != lastbyte && bitbuf->pos + throwaways < 8) {
-			bit = bb_read(bitbuf, 1);
-			free(bit);
+		int *count = bb_readbyte(bitbuf);
+		for (int i = 0; i < *count; i++) {
 			fwrite(&parent->ch, 1, 1, out);
 			fwrite(&parent->ch, 1, 1, out);
 		}
 		}
+		free(count);
 		return 0;
 		return 0;
 	}
 	}
-	#endif
 
 
 	while ((bit = bb_read(bitbuf, 1)) != NULL) {
 	while ((bit = bb_read(bitbuf, 1)) != NULL) {
 		printf("Reading bit %d\n", *bit);
 		printf("Reading bit %d\n", *bit);