19 enum rb_color { RED = 0U, BLACK = 1U };  enumerator
60 	return get_color(n) == RED;  in is_red()
153 /* The node at the top of the provided stack is red, and its parent is
154  * too.  Iteratively fix the tree so it becomes a valid red black tree
174 		 * parent is red, as red nodes cannot be the root  in fix_extra_red()
184 			set_color(grandparent, RED);  in fix_extra_red()
188 			/* We colored the grandparent red, which might  in fix_extra_red()
189 			 * have a red parent, so continue iterating  in fix_extra_red()
209 		set_color(stack[stacksz - 2], RED);  in fix_extra_red()
244 	set_color(node, RED);  in rb_insert()
261  * construction N must be black (because if it was red it would be
263  * the tree to preserve red/black rules.  The "null_node" pointer is
291 			set_color(parent, RED);  in fix_missing_black()
313 			set_color(sib, RED);  in fix_missing_black()
316 				 * coloring it red, then our parent  in fix_missing_black()
331 		/* We know sibling has at least one red child.  Fix it  in fix_missing_black()
333 		 * opposite side from N) is definitely red.  in fix_missing_black()
343 			set_color(sib, RED);  in fix_missing_black()
355 		/* Finally, the sibling must have a red child in the  in fix_missing_black()
484 			/* Red childless nodes can just be dropped */  in rb_remove()
490 		/* Check colors, if one was red (at least one must have been  in rb_remove()
493 		__ASSERT(is_black(node) || is_black(child), "both nodes red?!");  in rb_remove()