关于C#：Red Black Tree插入操作不断崩溃

Red Black Tree insertion operation keeps on crashing

我正在尝试实现Red Black Tree，但我自己无法对其进行编码，因此我在搜索它是用C编写的代码，并在github上找到了下面经过分析的代码，它非常简单明了，所以我做了稍作修改(只是重命名了几个变量并添加了一个菜单)，当我尝试运行它时，在尝试旋转树时遇到了一个问题。插入工作正常，直到树变得不平衡为止，特别是当叔叔变成"黑色"时，我的程序无法旋转树，并且崩溃了。因此，让我向您解释程序的流程：每当我们插入一个新节点时，都会调用一个名为insert的函数，在成功插入之后，它将调用另一个名为insertfixup的函数，该函数检查Red Black Tree的任何属性是否已经存在。如果叔叔是红色，那么它会根据问题节点的叔叔是红色还是黑色来旋转树，如果叔叔是红色，那么它可以正常工作，但是一旦叔叔变成黑色，它就会崩溃。有人可以检查一下我在下面提供的代码，并指出导致问题的确切原因吗，我怀疑它与指针有关，但无法弄清楚实际发生了什么，

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#include <stdio.h>
#include <stdlib.h>
#include <conio.h>

typedef struct RB_Tree {
struct RB_Tree *left, *right, *parent;
int info;
char color;
} node;
int count = 0;

void left_rotate(node **root, node *par) {
node *child = par->right;

par->right = child->left;

if (child->left!=NULL)
child->left->parent = par;

child->parent = par->parent;

if (par->parent == NULL)
(*root) = child;
else
if (par == par->parent->left)
par->parent->left = child;
else
par->parent->right = child;

child->left = par;
par->parent = child;
}

void right_rotate(node **root, node *par) {
node *child = par->left;

par->left = child->right;

if (child->right!=NULL)
child->right->parent = par;

child->parent = par->parent;

if (par->parent == NULL)
(*root) = child;
else
if (par = par->parent->left)
par->parent->left = child;
else
par->parent->right = child;

child->right = par;

par->parent = child;
}

void insertFixup(node **root, node *new) {
while (new != *root && new->parent->color == 'R') {
node *uncle;
//Find uncle and store uncle in"uncle" variable!

if (new->parent == new->parent->parent->left) {
uncle = new->parent->parent->right;
} else {
uncle = new->parent->parent->left;
}

if (uncle == NULL || uncle->color == 'B') {

if (new->parent == new->parent->parent->left && new == new->parent->left) {
printf("Inside ll case before rot");
char col = new->parent->color;
new->parent->color = new->parent->parent->color;
new->parent->parent->color = col;

right_rotate(root, new->parent->parent);
}

if (new->parent == new->parent->parent->right && new == new->parent->right) {
char col = new->parent->color;
new->parent->color = new->parent->parent->color;
new->parent->parent->color = col;
left_rotate(root, new->parent->parent);
}

if (new->parent == new->parent->parent->left && new == new->parent->right) {
char col = new->color;
new->color = new->parent->parent->color;
new->parent->parent->color = col;
//printf("\
Here we are in the left right!");
left_rotate(root, new->parent);

right_rotate(root, new->parent->parent);
}

if (new->parent == new->parent->parent->right && new == new->parent->left) {
char col = new->color;
new->color = new->parent->parent->color;
new->parent->parent->color = col;
right_rotate(root, new->parent);
left_rotate(root, new->parent->parent);
}
}

if (uncle) {
if (uncle->color == 'R') {
uncle->color = 'B';
new->parent->color = 'B';
new->parent->parent->color = 'R';
new = new->parent->parent;
}
}
}

(*root)->color = 'B';
}

void insert(node **root, int data) {
node *new = (node*)malloc(sizeof(node));
new->info = data;
new->parent = new->left = new->right = NULL;

if (*root == NULL) {
(*root) = new;
new->color = 'B';
} else {
node *par;
node *temp = (*root);

while (temp) {
par = temp;
if(new->info > temp->info)
temp = temp->right;
else
temp = temp->left;
}

new->parent = par;

if (par->info > new->info)
par->left = new;
else
par->right = new;

new->color = 'R';

insertFixup(root, new);
}
}

void main() {
int men, c, data;
node *root = NULL;

while (1) {
system("cls");
printf("1.) Insert\
");
printf("2.) exit\
");
printf("Enter your choice :");
scanf("%d", &men);
switch (men) {
case 1:
printf("Enter data :");
scanf("%d", &data);
insert(&root, data);
printf("%d successfully added!", data);
break;

case 2:
exit(0);

default:
printf("Invalid choice!");
while ((c = fgetc(stdin)) != '\
') {}
break;
}
getch();
}
}

相关讨论

主要问题

如前所述，第53行

else if(par = par->parent->left)

必须为

1	else if(par == par->parent->left)

insertFixup中的算法不正确，例如，先插入数据3，然后插入2，然后插入1，则会导致崩溃，我在互联网上看到一个相同的定义，可能您理解了，但不幸的是您的源已被窃听。该定义有效：

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void insertFixup(node **root, node *new) {
while (new != *root && new->parent->color == 'R') {
if (new->parent == new->parent->parent->left) {
if (new->parent->parent->right && new->parent->parent->right->color == 'R') {
new->parent->color = 'B';
new->parent->parent->right->color = 'B';
new->parent->parent->color = 'R';
new = new->parent->parent;
}
else {
if (new == new->parent->right) {
new = new->parent;
left_rotate(root, new);
}

new->parent->color = 'B';
new->parent->parent->color = 'R';
right_rotate(root, new->parent->parent);
}
}
else {
if (new->parent->parent->left && new->parent->parent->left->color == 'R') {
new->parent->color = 'B';
new->parent->parent->left->color = 'B';
new->parent->parent->color = 'R';
new = new->parent->parent;
}
else {
if (new == new->parent->left) {
new = new->parent;
right_rotate(root, new);
}

new->parent->color = 'B';
new->parent->parent->color = 'R';
left_rotate(root, new->parent->parent);
}
}
}

(*root)->color = 'B';
}

其他问题

* main *的签名错误，main返回一个int

如果用户未输入int，则您的scanf行160不会设置字符，并且您可以测试从未初始化的值。始终检查scanf的返回值，例如：与：

之后的代码兼容

1 2	if (scanf("%d", &men) != 1) men = -1;

在

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2
while ((c = fgetc(stdin)) != '\
') {}

设置了

c但从未使用过，例如，如果达到EOF，则最糟糕的情况是因为输入已重定向到文件上，因此您将循环播放而不会结束。例如do

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while ((c = fgetc(stdin)) != '\
')
if (c == EOF)
exit(0);

第10行：

1
int count = 0;

定义一个从未使用过的全局变量，您可以删除该行

要检查我的建议，我添加了该功能以显示树：

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void display(node * x){
if (x) {
display(x->left);
printf("%d", x->info);
display(x->right);
}
}

我修改了您的main以便能够显示树：

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int main()
{
int men, c, data;
node *root = NULL;

while (1) {
fputs("\
"
"1.) Insert\
"
"2.) exit\
"
"3.) Display tree\
"
"Enter your choice :",
stdout);
if (scanf("%d", &men) != 1)
men = -1;

switch (men) {
case 1:
printf("Enter data :");
scanf("%d", &data);
insert(&root, data);
printf("%d successfully added!", data);
break;

case 2:
exit(0);

case 3:
display(root);
putchar('\
');
break;

default:
puts("Invalid choice!");
while ((c = fgetc(stdin)) != '\
')
if (c == EOF)
exit(0);
break;
}
}
}

编译和执行：

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bruno@bruno-XPS-8300:/tmp/t$ gcc -Wall -Wextra t.c
bruno@bruno-XPS-8300:/tmp/t$ ./a.out

1.) Insert
2.) exit
3.) Display tree
Enter your choice : 1
Enter data : 3
3 successfully added!
1.) Insert
2.) exit
3.) Display tree
Enter your choice : 1
Enter data : 2
2 successfully added!
1.) Insert
2.) exit
3.) Display tree
Enter your choice : 1
Enter data : 1
1 successfully added!
1.) Insert
2.) exit
3.) Display tree
Enter your choice : 3
1 2 3

1.) Insert
2.) exit
3.) Display tree
Enter your choice : 1
Enter data : 5
5 successfully added!
1.) Insert
2.) exit
3.) Display tree
Enter your choice : 1
Enter data : 4
4 successfully added!
1.) Insert
2.) exit
3.) Display tree
Enter your choice : 3
1 2 3 4 5

1.) Insert
2.) exit
3.) Display tree
Enter your choice : 2
bruno@bruno-XPS-8300:/tmp/t$