c-基数树实现中的无限循环问题

Question

我的基数树实现有问题。我的想法是创建第一个节点，然后输入一些二进制数字。二进制数确定是创建左节点(0)还是创建右节点(1)。一旦我到达二进制数的末尾，我就会将一个节点设置为"活动"。

然后我在树中搜索以找到一个活动节点，并通过检查到达活动节点的方向再次输出原始二进制数。

这是完整的代码：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef int bool;
enum { false, true };
typedef struct radixNode {
bool active;
struct radixNode * pnt;
struct radixNode * l;
struct radixNode * r;
} node;
void insert(node *root, char * B) {
printf("String: %sn", B);
printf("1st: %c", B[0]);
printf("nn", B);
// digit is zero so we go left
if (B[0] == '0') {
printf("till here if");
// left child doesn't exist, create it
if (root->l == NULL) {
root->l = malloc(sizeof(node));
/* if the next index in the string does NOT contain a 1 or 0,
the current index is the last index and the node is activated */
if (B[1] == 1 || B[1] == 0)
root->l->active = false;
else
root->l->active = true;
root->l->pnt = root;
root->l->l = NULL;
root->l->r = NULL;
insert(root->l,B++); // B++ removes the first digit of the string
}
// left child exists, traverse
else {
insert(root->l,B++);
}
}
// digit is one, go right
else if (B[0] == '1') {
printf("first was 1n");
// right child doesn't exist, create it
if (root->r == NULL) {
printf("if triggeredn");
root->r = malloc(sizeof(node));
/* if the next index in the string does NOT contain a 1 or 0,
the current index is the last index and the node is activated */
if (B[1] == 1 || B[1] == 0)
root->r->active = false;
else
root->r->active = true;
root->r->pnt = root;
root->r->l = NULL;
root->r->r = NULL;
insert(root->r,B++);
}
// left child exists, traverse
else {
printf("else triggeredn");
insert(root->r,B++);
}
}
}
node * printTreeMin(node *root) {
char C[10];
/* goes left until it can't, appends 0 to string
till it can't. if node is active, print the string */
while (root->l != NULL) {
C[strlen(C)] = '0';
if (root->active == true)
printf("%sn",C);
root = root->l;
}
return root;
}
// prints the next smallest binary number in the tree, returns the node it printed
node * printNextSmallest(node * root) {
char C[10];
// if right child exists, go there and find lowest node (after if same deal as printTreeMin() )
if (root->r != NULL) {
C[strlen(C)] = '1';
if (root->active == true)
printf("%sn",C);
root = root->r;
while (root->l != NULL) {
C[strlen(C)] = '0';
if (root->active == true)
printf("%sn",C);
root = root->l;
}
return root;
}
node * temp = root->pnt;
while (temp != NULL && root == temp->r) {
root = temp;
temp = temp->pnt;
}
return temp;
}
void printRadixTree(node *root) {
root = printTreeMin(root);
while (printNextSmallest(root) != NULL)
root = printNextSmallest(root);
}
void test() {
node * tree = malloc(sizeof(node));
tree->l = NULL;
tree->r = NULL;
// a)
insert(tree,"101000");
insert(tree,"10100");
insert(tree,"10110");
insert(tree,"101");
insert(tree,"1111");
// b)
printRadixTree(tree);
}
int main() {
test();
}

这是输出：

if triggered
String: 101000
1st: 1
first was 1
if triggered
String: 101000
1st: 1
first was 1
if triggered
String: 101000
1st: 1

很明显，我在insert()函数的递归中遇到了问题，但考虑到我在递归时删除了二进制数字字符串的第一个字符，我不明白它是如何无限运行的。

Answer 1

无限递归的原因是您选择了自动递增运算符。你想要前缀，而不是后缀形式。

insert(..., B++)

在调用insert后递增指针(去掉第一个字符)。

相反，呼叫应该是

insert (..., ++B)

你的active标志也有问题，这就是你的罪魁祸首

if (B[1] == 1 || B[1] == 0)

我想你是指

if (B[1] == '1' || B[1] == '0')

第一种形式是检查二进制零或一，而不是ASCII字符。

这样做的结果是，对于大多数节点，active标志可能设置不正确。我预计在遍历树时会出现问题。事实上，只有当您查看字符串中的最后一个'0'或'1'时，active才会设置为false(因为此时的B[1]将是终止的'')。

此外，对于回避例程，使基本情况明确而不是隐含总是一个好主意。因此，insert中的第一个代码块可能是

if (B[0] != '1' && B[0] != `0`)
return;

然后你可以用一个简单的else替换else if

if (B[0] == '0')
{
// ... go left
}
else
{
// ... go right
}