一步一圖一程式碼，一定要讓你真正徹底明白紅黑樹 （July演算法！！！）
阿新 • • 發佈：2019-01-08
情況5: 兄弟S為黑色，S 的左兒子是紅色，S 的右兒子是黑色，而N是它父親的左兒子。
//此種情況，最後轉化到下面的情況6。
[對應我第二篇文章中，情況3：x的兄弟w是黑色的，w的左孩子是紅色，w的右孩子是黑色。]
void delete_case5(struct node *n)
{
struct node *s = sibling(n);
if (s->color == BLACK)
if ((n == n->parent->left) &&
(s->right->color == BLACK) &&
(s->left->color == RED)) {
// this last test is trivial too due to cases 2-4.
s->color = RED;
s->left->color = BLACK;
rotate_right(s);
} else if ((n == n->parent->right) &&
(s->left->color == BLACK) &&
(s->right->color == RED)) {
// this last test is trivial too due to cases 2-4.
s->color = RED;
s->right->color = BLACK;
rotate_left(s);
}
}
delete_case6(n); //轉到情況6。
//此種情況，最後轉化到下面的情況6。
[對應我第二篇文章中，情況3：x的兄弟w是黑色的，w的左孩子是紅色，w的右孩子是黑色。]
void delete_case5(struct node *n)
{
struct node *s = sibling(n);
if (s->color == BLACK)
if ((n == n->parent->left) &&
(s->right->color == BLACK) &&
(s->left->color == RED)) {
// this last test is trivial too due to cases 2-4.
s->color = RED;
s->left->color = BLACK;
rotate_right(s);
} else if ((n == n->parent->right) &&
(s->left->color == BLACK) &&
(s->right->color == RED)) {
// this last test is trivial too due to cases 2-4.
s->color = RED;
s->right->color = BLACK;
rotate_left(s);
}
}
delete_case6(n); //轉到情況6。