C language to achieve the red black tree instance code
- 2020-04-02 02:05:43
- OfStack
Because the redblack tree is fun to look at, take advantage of the weekend to play with it. The operation of the red-black tree is mainly insert and delete, and delete needs to consider more cases. Specific operation is not wordy here, baidu library there is a better article, has said very clear.
Some people may feel that some situations are not considered when they look at the specific operation (if they do not feel this way, they may not think carefully at all). But the "missing" cases, if any, would violate those rules.
When I wrote the code, I made many mistakes because I didn't think much about the situation. There were too many details. At the beginning, there was no pointer to the parent node in rb_node, so I wrote it fast and then added it. The specific meaning of the code can be seen in combination with the article and comments (which are easy to understand). The following code may not be considered in the details, welcome to the brick.
#include <stdio.h>
#include <stdlib.h>
const int RED = 0;
const int BLACK = 1;
struct rb_node{
rb_node* lchild, *rchild, *parent;
int key, colour;
};
rb_node* root;
rb_node* get_node(rb_node* parent, int key);
void rb_insert(int key);
rb_node* rb_search(int key);
void rb_delete(int key);
rb_node* clock_wise(rb_node* node);
rb_node* counter_clock_wise(rb_node* node);
void show_rb_tree(rb_node* node);
rb_node* get_node(rb_node* parent, int key){
rb_node *tmp = (rb_node*)malloc(sizeof(rb_node));
tmp->key = key;
tmp->colour = RED;
tmp->parent = parent;
tmp->lchild = tmp->rchild = NULL;
return tmp;
}
rb_node* clock_wise(rb_node* node){
if(node == NULL || node->lchild == NULL)
return NULL;
rb_node *rb_1=node, *rb_2=node->lchild, *rb_3=node->lchild->rchild;
if(rb_1->parent != NULL){
if(rb_1->parent->lchild == rb_1)
rb_1->parent->lchild = rb_2;
else
rb_1->parent->rchild = rb_2;
}else if(rb_1 == root){
root = rb_2;
}
rb_2->parent = rb_1->parent;
rb_1->parent = rb_2;
rb_2->rchild = rb_1;
rb_1->lchild = rb_3;
if(rb_3 != NULL)rb_3->parent = rb_1;
return rb_2;
}
rb_node* counter_clock_wise(rb_node* node){
if(node == NULL || node->rchild == NULL)
return NULL;
rb_node *rb_1=node, *rb_2=node->rchild, *rb_3=node->rchild->lchild;
if(rb_1->parent != NULL){
if(rb_1->parent->lchild == rb_1)
rb_1->parent->lchild = rb_2;
else
rb_1->parent->rchild = rb_2;
}
else if(rb_1 == root){
root = rb_2;
}
rb_2->parent = rb_1->parent;
rb_1->parent = rb_2;
rb_2->lchild = rb_1;
rb_1->rchild = rb_3;
if(rb_3 != NULL)rb_3->parent = rb_1;
return rb_2;
}
rb_node* rb_search(int key){
rb_node *p = root;
while(p != NULL){
if(key < p->key)
p = p->lchild;
else if(key > p->key)
p = p->rchild;
else
break;
}
return p;
}
void rb_insert(int key){
rb_node *p=root, *q=NULL;
if(root == NULL){
root = get_node(NULL, key);
root->colour = BLACK;
return;
}
while(p != NULL){
q = p;
if(key < p->key)
p = p->lchild;
else if(key > p->key)
p = p->rchild;
else return;
}
if(key < q->key)
q->lchild = get_node(q, key);
else
q->rchild = get_node(q, key);
while(q != NULL && q->colour == RED){
p = q->parent;//p won't null, or BUG.
if(p->lchild == q){
if(q->rchild != NULL && q->rchild->colour == RED)
counter_clock_wise(q);
q = clock_wise(p);
q->lchild->colour = BLACK;
}
else{
if(q->lchild != NULL && q->lchild->colour == RED)
clock_wise(q);
q = counter_clock_wise(p);
q->rchild->colour = BLACK;
}
q = q->parent;
}
root->colour = BLACK;
}
void show_rb_tree(rb_node* node){
if(node == NULL)
return;
printf("(%d,%d)n", node->key, node->colour);
if(node->lchild != NULL){
printf("[-1]n");
show_rb_tree(node->lchild);
}
if(node->rchild != NULL){
printf("[1]n");
show_rb_tree(node->rchild);
}
printf("[0]n");
}
void rb_delete(int key){
rb_node *v = rb_search(key), *u, *p, *c, *b;
int tmp;
if(v == NULL) return;
u = v;
if(v->lchild != NULL && v->rchild != NULL){
u = v->rchild;
while(u->lchild != NULL){
u = u->lchild;
}
tmp = u->key;
u->key = v->key;
v->key = tmp;
}
//u is the node to free.
if(u->lchild != NULL)
c = u->lchild;
else
c = u->rchild;
p = u->parent;
if(p != NULL){
//remove u from rb_tree.
if(p->lchild == u)
p->lchild = c;
else
p->rchild = c;
}
else{
//u is root.
root = c;
free((void*)u);
return;
}
//u is not root and u is RED, this will not unbalance.
if(u->colour == RED){
free((void*)u);
return;
}
free((void*)u);
u = c;
//u is the first node to balance.
while(u != root){
if(u != NULL && u->colour == RED){
//if u is RED, change it to BLACK can finsh.
u->colour = BLACK;
return;
}
if(u == p->lchild)
b = p->rchild;
else
b = p->lchild;
printf("%dn", b->key);
//b is borther of u. b can't be null, or the rb_tree is must not balance.
if(b->colour == BLACK){
//If b's son is RED, rotate the node.
if(b->lchild != NULL && b->lchild->colour == RED){
if(u == p->lchild){
b = clock_wise(b);
b->colour = BLACK;
b->rchild->colour = RED;
p = counter_clock_wise(p);
p->colour = p->lchild->colour;
p->lchild->colour = BLACK;
p->rchild->colour = BLACK;
}
else{
p = clock_wise(p);
p->colour = p->rchild->colour;
p->rchild->colour = BLACK;
p->lchild->colour = BLACK;
}
return;
}
else if(b->rchild != NULL && b->rchild->colour == RED){
if(u == p->rchild){
b = counter_clock_wise(b);
b->colour = BLACK;
b->lchild->colour = RED;
p = clock_wise(p);
p->colour = p->rchild->colour;
p->rchild->colour = BLACK;
p->lchild->colour = BLACK;
}
else{
p = counter_clock_wise(p);
p->colour = p->lchild->colour;
p->lchild->colour = BLACK;
p->rchild->colour = BLACK;
}
return;
}
else{//if b's sons are BLACK, make b RED and move up u.
b->colour = RED;
u = p;
p = u->parent;
continue;
}
}
else{
if(u == p->lchild){
p = counter_clock_wise(p);
p->colour = BLACK;
p->lchild->colour = RED;
p = p->lchild;
}
else{
p = clock_wise(p);
p->colour = BLACK;
p->rchild->colour = RED;
p = p->rchild;
}
}
}
root->colour = BLACK;
}
int main(){
int i;
root = NULL;
for(i = 1; i <= 10; i++){
rb_insert(i);
}
rb_delete(9);
rb_delete(3);
rb_delete(7);
show_rb_tree(root);
printf("n");
return 0;
}