C 红黑树插入码
我正试图为自己的学习编写一个红黑树的实现，我已经盯着它看了几天了 有人能帮我弄清楚如何使双旋转箱正常工作吗？如果你在浏览这些片段时发现了其他糟糕的东西，请随意让我觉得自己像个白痴 谢谢你的帮助 再平衡功能C 红黑树插入码,c,red-black-tree,rebalancing,red-black-tree-insertion,C,Red Black Tree,Rebalancing,Red Black Tree Insertion,我正试图为自己的学习编写一个红黑树的实现，我已经盯着它看了几天了 有人能帮我弄清楚如何使双旋转箱正常工作吗？如果你在浏览这些片段时发现了其他糟糕的东西，请随意让我觉得自己像个白痴 谢谢你的帮助 再平衡功能 int impl_rbtree_rebalance (xs_rbtree_node *node) { check_mem(node); xs_rbtree_node *p = node->parent; if (!p) return 0; if (p-
int impl_rbtree_rebalance (xs_rbtree_node *node)
{
check_mem(node);
xs_rbtree_node *p = node->parent;
if (!p) return 0;
if (p->color == BLACK) return 0;
xs_rbtree_node *gp = p->parent;
if (!gp) return 0;
xs_rbtree_node *uncle;
int is_parent_left_child;
/* check if parent is left or right child */
if (p == gp->left) {
uncle = gp->right;
is_parent_left_child = 1;
} else {
uncle = gp->left;
is_parent_left_child = 0;
}
if (uncle && uncle->color == RED) {
p->color = uncle->color = BLACK;
gp->color = RED;
} else { /* uncle is black */
if (gp->parent == NULL) return 0;
if (node == p->left) {
if (is_parent_left_child == 0) {
/* Double rotation logic */
} else {/* parent is left child */
gp->color = RED;
gp->left->color = BLACK;
impl_rbtree_rr(&gp);
}
} else { /* node is right child */
if (is_parent_left_child == 1) {
/* Double rotation logic */
} else { /* parent is right child */
gp->color = RED;
gp->right->color = BLACK;
impl_rbtree_rl(&gp);
}
}
}
return 0;
}
xs_rbtree_node *impl_rbtree_node_alloc (xs_rbtree_node *parent, void *val)
{
xs_rbtree_node *n = malloc(sizeof(xs_rbtree_node));
n->parent = parent;
n->val = val;
n->left = n->right = NULL;
n->color = (parent == NULL) ? BLACK : RED;
return n;
}
void rbtree_insert (xs_rbtree_ *tree, void *val)
{
check_mem(tree);
check_mem(val);
tree->root = impl_rbtree_insert(tree->cmp, NULL, tree->root, val);
tree->root->color = BLACK;
}
xs_rbtree_node *impl_rbtree_insert (xs_tree_cmp cmp, xs_rbtree_node *parent, xs_rbtree_node *node, void *val)
{
check_mem(cmp);
check_mem(val);
if (!node) {
node = impl_rbtree_node_alloc(parent, val);
} else if (cmp(node->val, val) < 0) {
/* node->val < val */
check_mem(node);
node->right = impl_rbtree_insert(cmp, node, node->right, val);
impl_rbtree_rebalance(node->right);
} else if (cmp(node->val, val) > 0) {
/* node->val > val */
check_mem(node);
node->left = impl_rbtree_insert(cmp, node, node->left, val);
impl_rbtree_rebalance(node->left);
}
/* ignoring values that are equal */
return node;
}
#include <xs/base/bstree.h>
void impl_tree_rr(xs_tree_node **node)
{
check_mem(*node);
check_mem((*node)->left);
xs_tree_node *k1, *k2;
k2 = *node;
k1 = k2->left;
k2->left = k1->right;
k1->right = k2;
k1->parent = k2->parent;
k2->parent = k1;
*node = k1;
}
void impl_tree_rl(xs_tree_node **node)
{
check_mem(*node);
check_mem((*node)->right);
xs_tree_node *k1, *k2;
k2 = *node;
k1 = k2->right;
k2->right = k1->left;
k1->left = k2;
k1->parent = k2->parent;
k2->parent = k1;
*node = k1;
}
void impl_tree_drr(xs_tree_node **node)
{
check_mem(*node);
impl_tree_rl(&((*node)->left));
impl_tree_rr(node);
}
void impl_tree_drl(xs_tree_node **node)
{
check_mem(*node);
impl_tree_rr(&((*node)->right));
impl_tree_rl(node);
}
typedef struct xs_rbtree_node xs_rbtree_node;
typedef struct xs_rbtree_ xs_rbtree_;
typedef enum { RED, BLACK } impl_rbtree_color;
struct xs_rbtree_node {
xs_rbtree_node *left;
xs_rbtree_node *right;
xs_rbtree_node *parent;
void *val;
impl_rbtree_color color;
};
struct xs_rbtree_ {
xs_rbtree_node *root;
xs_tree_cmp cmp;
};
与AVL树相比，红黑树中的双旋转有点棘手，原因是RB树中存在父指针。 当我们旋转树时，我们也需要调整父指针。 我将用一个例子来说明这一点。下面的树是初始树 这是右旋的 我们在那里观察到两件事。 1.根变了。 2.3个节点的父指针已更改（10、20、15），即新根、旧根和新根的右子节点。这对我来说是真实的 好的，轮换。这是为了确保 旋转发生后，遍历保持不变 现在看看我们如何一步一步地旋转
public Node rotateRight() {
if(root.left == null) {
System.out.println("Cannot be rotated Right");
} else {
Node newRoot = root.left; // Assign new root.
Node orphaned = newRoot.right; // Store ref to the right child of new root.
newRoot.parent = root.parent; // new root parent point to old root parent.
root.parent = newRoot; // new root becomes parent of old root.
newRoot.right = root; // old root becomes right child of new root
root.left = orphaned;
if (orphaned != null) {
orphaned.parent = root;
}
root = newRoot;
}
return root;
}
a. We need to see if the grandpa is right or left child of its
parent and assign the root accordingly.
b. If the grandpa has no parent means we are doing rotation at the level of tree's root, then we reassign the root of the tree.
temp is the new node which is added.
// if new node is the right child AND parent is right child
else if(temp.parent.right == temp && grandPa!=null && grandPa.right == temp.parent) {
// If becomes a right-right case
Boolean isLeft = isLeftChild(grandPa);
if(isLeft == null) {
// Grandpa has no parent, reassign root.
root = rotateLeft(grandPa,true);
}
else if(isLeft) {
grandPa.parent.left = rotateLeft(grandPa,true);
} else {
grandPa.parent.right = rotateLeft(grandPa,true);
}
}
我认为你应该在论坛上发布这类问题的地方是谢谢你的评论。我也会把它贴在那里。我只是想，因为这是一个实现问题，而不是关于代码质量的问题，所以它更适合这样做@洛托洛
a. We need to see if the grandpa is right or left child of its
parent and assign the root accordingly.
b. If the grandpa has no parent means we are doing rotation at the level of tree's root, then we reassign the root of the tree.
temp is the new node which is added.
// if new node is the right child AND parent is right child
else if(temp.parent.right == temp && grandPa!=null && grandPa.right == temp.parent) {
// If becomes a right-right case
Boolean isLeft = isLeftChild(grandPa);
if(isLeft == null) {
// Grandpa has no parent, reassign root.
root = rotateLeft(grandPa,true);
}
else if(isLeft) {
grandPa.parent.left = rotateLeft(grandPa,true);
} else {
grandPa.parent.right = rotateLeft(grandPa,true);
}
}