C 二叉树,内存分配
所以我尝试用C语言用结构和jazz来构建一个二叉树。到目前为止,我得到了以下信息:C 二叉树,内存分配,c,memory-management,struct,binary-search-tree,C,Memory Management,Struct,Binary Search Tree,所以我尝试用C语言用结构和jazz来构建一个二叉树。到目前为止,我得到了以下信息: #include <stdio.h> #include <string.h> #include <stdarg.h> #include <stdlib.h> #include <math.h> #include <assert.h> #include <signal.h> typedef struct treeNode {
#include <stdio.h>
#include <string.h>
#include <stdarg.h>
#include <stdlib.h>
#include <math.h>
#include <assert.h>
#include <signal.h>
typedef struct treeNode {
int value;
struct treeNode *left;
struct treeNode *right;
} node;
node *top = NULL;
static node *addNode(int e, node *n);
static void printTree(node *n);
static node *addNode(int e, node *n) {
if(top == NULL) {
top = malloc(sizeof(node));
top->value = e;
top->left = NULL;
top->right = NULL;
return top;
} else if (n != NULL) {
if(e <= n->value) {
if(n->left == NULL) {
n->left = addNode(e, n->left);
} else {
(void) addNode(e, n->left);
}
} else {
if(n->right == NULL) {
n->right = addNode(e, n->right);
} else {
(void) addNode(e, n->right);
}
}
} else if(n == NULL) {
n = malloc(sizeof(node));
n->value = e;
n->left = NULL;
n->right = NULL;
return n;
}
return n;
}
static void printTree(node *n) {
if(n != NULL) {
fprintf(stdout, "%d, ", n->value);
if(n->left != NULL) {
printTree(n->left);
}
if(n->right != NULL) {
printTree(n->right);
}
}
}
int main(int argc, char **argv) {
addNode(1, top);
addNode(2, top);
addNode(0, top);
addNode(3, top);
printTree(top);
return 0;
}
这根本不起作用-因此没有创建新节点。现在的情况是,它正在按照它必须的方式工作——我真的不明白它为什么会这样做。当我第一次调用addNode函数时,不管怎样,我都会将顶部指针作为参数提供给它,因此它应该检查它是否为null(此时应该为null),并且应该为它分配一些内存。然而,这并没有发生。我不明白为什么。您需要将第一个节点分配给
top
int main(int argc, char **argv) {
top = addNode(1, top);
addNode(2, top);
addNode(0, top);
addNode(3, top);
printTree(top);
return 0;
}
对于原始的
if
子句,top
未得到更新。您刚才在这里返回了几点。首先让我们从细节开始:您的printTree()
太复杂了。当您在输入函数时检查空值时,无需再次检查左/右:
void printTree(node *n) {
if (n == NULL) { return; } // just leave
printTree(n->left);
printf("%d\n", n->value);
printTree(n->right);
}
请注意,打印值的顺序(左和右)将改变输出。在本例中,它将从低值打印到高值
现在,对于addNode()
来说,不需要全局top
值。将top
赋给函数,并获取带有返回值的新函数(top=addNode(val,top);
。在混合n
和top
访问时,使用此全局值只会导致错误代码。
而且功能可以更简单:
node *addNode(int val, node *n) {
if (n == NULL) { // just allocate a new one
n = malloc(sizeof(node)); // add check for failure here, of course
n->left = n->right = NULL; // it is a leaf, no child
n->value = val;
return(n); // return the node for caller
}
// ok, we are on a node, check on which side we need to insert
if (val < n->value) {
// to the left: it may be replaced (if created)
n->left = addNode(val, n->left);
} else {
// same, for right
n->right = addNode(val, n->right);
}
// return the node so that top=addNode(val,top) always works
return(n);
}
不,
top
是在addNode
中设置的。看一下对malloc
的调用top
是全局的。所以这不会有任何区别。这段代码是为了让他最初的尝试在他将if n==NULL
放在第一位的地方工作,而根本没有设置top
。哦,好的,对不起,我误解了:(*n)->左
(右和值相同)
node *addNode(int val, node *n) {
if (n == NULL) { // just allocate a new one
n = malloc(sizeof(node)); // add check for failure here, of course
n->left = n->right = NULL; // it is a leaf, no child
n->value = val;
return(n); // return the node for caller
}
// ok, we are on a node, check on which side we need to insert
if (val < n->value) {
// to the left: it may be replaced (if created)
n->left = addNode(val, n->left);
} else {
// same, for right
n->right = addNode(val, n->right);
}
// return the node so that top=addNode(val,top) always works
return(n);
}
void addNode(int val, node **n) {
if (*n == NULL) { // just allocate a new one
*n = malloc(sizeof(node)); // add check for failure here, of course
(*n)->left = (*n)->right = NULL; // it is a leaf, no child
(*n)->value = val;
return; // return, the node 'n' is still modified
}
// ok, we are on a node, check on which side we need to insert
if (val < (*n)->value) {
// to the left: it may be replaced (if created)
(*n)->left = addNode(val, (*n)->left);
} else {
// same, for right
(*n)->right = addNode(val, (*n)->right);
}
// leave function (return not needed), node unchanged
return;
}
int main(void) {
node *top = NULL; // must be initialized
addNode(3, &top);
addNode(1, &top);
addNode(5, &top);
printNode(top);
(…)