Question

我试图弄清楚如何在二叉树中插入大量数字。程序通过询问一个数字开始，并将0插入用户输入的任何数字。我的程序工作但它开始崩溃大约40,000我得到错误：流程返回-1073741571（0XC00000FD）

我是C和内存管理的新手，但我相信问题就在那里。任何建议？

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

typedef struct node BSTREE;

struct node
{
    int data;
    struct node * left;
    struct node * right;
};

void insert(BSTREE ** root, int number);

 int main()
{

    BSTREE *root = malloc(sizeof *root);
    BSTREE *tmp = malloc(sizeof *root);
    root = NULL;
    int x;
    int input;

    FILE *fp;
    fp=fopen("c:\\test2.txt", "w");


    printf( "Please enter a number: " );
    scanf( "%d", &input );

    for ( x = 0; x < input; x++ )
    {
    insert(&root, x);
    }
    printf("%d", x);
    free(root);

    fclose(fp);

    }

void insert(BSTREE ** root, long int number)
{

    BSTREE *temp = NULL;
    if(!(*root))
    {
        temp = (BSTREE *)malloc(sizeof(BSTREE));
        temp->left = temp->right = NULL;
        temp->data = number;
        *root = temp;
        return;
    }

    if(number < (*root)->data)
    {
        insert(&(*root)->left, number);
        free(root);
    }
    else if(number > (*root)->data)
    {
        insert(&(*root)->right, number);
        free(root);
    }
 free(root);
   }

Answer 1

每次调用insert时，您都会调用free(root)两次。在insert返回后立即执行。然后立即再次在同一节点上。并且您似乎总是自由地使用您刚刚插入的节点。因此，你的记忆可能已经腐败，恰好在小树上工作。随着树的变大，内存开始重新分配，树腐败变得更加明显。

这是一个更简单的解决方案。它更详细，但在一个函数中处理malloc和free，而核心递归插入在另一个函数中完成。希望这会有所帮助。

extern int insert_node(BSTREE* root, BSTREE* node);

void insert(BSTREE** root, int number)
{
    /* allocate a node for our number to be inserted */
    BSTREE* node = malloc(sizeof(BSTREE));
    node->data = number;
    node->left = NULL;
    node->right = NULL;
    int insert_result;

    if (*root == NULL)
    {
        /* handle the empty tree case */
        *root = node;
        return;
    }

    insert_result = insert_node(*root, node);

    if (!insert_result)
    {
        /* insert_node returned false.
           This means it already had a duplicate of the node 
           in the tree. Just free it since it did not get inserted
        */
        free(node);
        node = NULL;
    }

}

int insert_node(BSTREE* root, BSTREE* node)
{
    int result = 1; /* success */

    if (node->data < root->data)
    {
        /* traverse left */
        if (node->left != NULL)
        {
            result = insert_node(root->left, node);
        }
        else
        {
            root->left = node;
        }
    }
    else if (node->data > root->data)
    {
        /* traverse right */
        if (node->right != NULL)
        {
            result = insert_node(root->right, node);
        }
        else
        {
            root->right = node;
        }
    }
    else
    {
        /* else node is equal to root - nothing to do */
        /* return 0 to indicate the node should be free'd by the caller*/
        result = 0;
    }

    return result;
}

循环插入二叉树

1 个答案: