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我正在尝试自己从头开始构建 BST 以完成一项任务，而我所拥有的只是几个月的 C 语言经验。在添加新节点时遇到了困难，显然每次都更换了 Root。这里有几个片段。

结构：

typedef struct node
{ 
    int Num;
    struct node * left;     /* Points to the left child*/
    struct node * right;    /*Points to the right child*/
    struct node *parent;

}Node;

typedef struct tree
{   
    Node * root; /*Points to the root*/
    int height;
}Tree;

职能：

void AddNode(int num, Tree * tree){
    Node NewNode;
    NewNode.Num=num;
    NewNode.left=NULL;
    NewNode.right=NULL;
    NewNode.parent=NULL;

    Compare(tree,&NewNode,tree->root);
}

void Compare(Tree * tree,Node * newNode,Node * oldNode)
{
    if (oldNode == NULL)
        {
            oldNode=newNode;
            printf("hi1");
        }
    else if (newNode->Num<oldNode->Num)
        {
        printf("hi2");
            if (oldNode->left==NULL)
                {
                    oldNode->left=newNode;
                    oldNode->left->parent=oldNode;
                }
            else
                Compare(tree,newNode,oldNode->left);
        }
    else if (newNode->Num>oldNode->Num)
        {
            printf("hi3");
            if (oldNode->right==NULL)
                {
                    oldNode->right=newNode;
                    oldNode->right->parent=oldNode;
                }
            else
                Compare(tree,newNode,oldNode->right);
        }
}

我有一个带有一些数字的数组。一开始我初始化了树。printfs 用于调试。输出是：

Tree Initialized
hi1hi1hi1hi1

这必须意味着每次都在更换根。我试着带走

if (oldNode == NULL)
            {
                oldNode=newNode;
                printf("hi1");
            }

并在调用 AddNode 方法之前将根设为 array[0]，但是程序停止执行。有任何想法吗？

4

2 回答 2

3

一个主要问题在于AddNode()：

void AddNode(int num, Tree * tree)
{
    Node NewNode;
    NewNode.Num=num;
    NewNode.left=NULL;
    NewNode.right=NULL;
    NewNode.parent=NULL;
    Compare(tree,&NewNode,tree->root);
}

当您从函数返回时，存储空间NewNode会在下一次调用时重用。您需要动态分配新节点：

void AddNode(int num, Tree * tree)
{
    Node *NewNode = malloc(sizeof(*NewNode));
    if (NewNode == 0)
        abort();
    NewNode->Num=num;
    NewNode->left=NULL;
    NewNode->right=NULL;
    NewNode->parent=NULL;
    Compare(tree, NewNode, tree->root);
}

还有其他问题——仍在解决这些问题。如果您展示了用于 (a) 填充 BST 和 (b) 用于打印 BST 的内容和 (c) 释放 BST 使用的存储的代码，将会有所帮助。如果没有 (b) 项，则无法调试自己的结构。

一个SSCCE

#include <assert.h>
#include <inttypes.h>
#include <stdio.h>
#include <stdlib.h>

typedef struct node
{ 
    int Num;
    struct node *left;     /* Points to the left child */
    struct node *right;    /* Points to the right child */
    struct node *parent;
} Node;

typedef struct tree
{   
    Node *root;
    int   height;
} Tree;

static void Compare(Node *newNode, Node **oldNode)
{
    if (*oldNode == NULL)
    {
        *oldNode = newNode;
        printf("hi1\n");
    }
    else if (newNode->Num < (*oldNode)->Num)
    {
        printf("hi2\n");
        if ((*oldNode)->left==NULL)
        {
            (*oldNode)->left=newNode;
            (*oldNode)->left->parent = *oldNode;
        }
        else
            Compare(newNode, &(*oldNode)->left);
    }
    else if (newNode->Num > (*oldNode)->Num)
    {
        printf("hi3\n");
        if ((*oldNode)->right==NULL)
        {
            (*oldNode)->right=newNode;
            (*oldNode)->right->parent = *oldNode;
        }
        else
            Compare(newNode, &(*oldNode)->right);
    }
} 

static void AddNode(int num, Tree *tree)
{
    assert(tree != 0);
    Node *NewNode = malloc(sizeof(*NewNode));
    if (NewNode == 0)
        abort();
    NewNode->Num    = num;
    NewNode->left   = NULL;
    NewNode->right  = NULL;
    NewNode->parent = NULL;
    Compare(NewNode, &tree->root);
}

static void DumpNode(const Node *node)
{
    assert(node != 0);
    printf("Number: %4d - Node: 0x%.9" PRIXPTR " (parent 0x%.9" PRIXPTR ", left 0x%.9" PRIXPTR ", right 0x%.9" PRIXPTR ")\n",
            node->Num, (uintptr_t)node, (uintptr_t)node->parent, (uintptr_t)node->left, (uintptr_t)node->right);
    if (node->left != 0)
        DumpNode(node->left);
    if (node->right != 0)
        DumpNode(node->right);
}

static void DumpTree(const char *tag, const Tree *tree)
{
    assert(tag != 0 && tree != 0);
    printf("BST Dump: %s\n", tag);

    printf("Tree: 0x%.9" PRIXPTR " (root 0x%.9" PRIXPTR ", height %d)\n",
           (uintptr_t)tree, (uintptr_t)tree->root, tree->height);
    if (tree->root != 0)
        DumpNode(tree->root);
}

static void FreeNode(Node *node)
{
    assert(node != 0);
    if (node->left != 0)
    {
        FreeNode(node->left);
        node->left = 0;
    }
    if (node->right != 0)
    {
        FreeNode(node->right);
        node->right = 0;
    }
    node->parent = 0;
    node->Num    = 0;
    free(node);
}


static void FreeTree(Tree *tree)
{
    assert(tree != 0);
    if (tree->root != 0)
        FreeNode(tree->root);
}

int main(void)
{
    char buffer[32];
    int list[] = { 19, 7, 12, 15, 21, 8, 16, 1, 3, 5 };
    enum { LIST_SIZE = sizeof(list) / sizeof(list[0]) };

    Tree tree = { 0, 0 };

    DumpTree(buffer, &tree);

    for (int i = 0; i < LIST_SIZE; i++)
    {
        AddNode(list[i], &tree);
        snprintf(buffer, sizeof(buffer), "After %2d: value %2d", i, list[i]);
        DumpTree(buffer, &tree);
    }

    FreeTree(&tree);
    return 0;
}

主要变化是该Compare()函数不需要tree参数，并且确实需要Node **oldNode参数。其余部分都有相应的变化Compare()。中的代码AddNode()几乎与最初建议的一样；当然，区别在于对 的调用Compare()。

添加了DumpTree()和FreeTree()函数。0x%.9X在我的 64 位机器上，这些格式是统一的，堆地址在 0x100000000 及以上。有了这些和简单main()的，我运行的输出valgrind是：

==48994== Memcheck, a memory error detector
==48994== Copyright (C) 2002-2012, and GNU GPL'd, by Julian Seward et al.
==48994== Using Valgrind-3.8.1 and LibVEX; rerun with -h for copyright info
==48994== Command: ./bst
==48994== 
==48994== Conditional jump or move depends on uninitialised value(s)
==48994==    at 0xE567: strlen (mc_replace_strmem.c:407)
==48994==    by 0x1778C2: __vfprintf (in /usr/lib/system/libsystem_c.dylib)
==48994==    by 0x17618D: vfprintf_l (in /usr/lib/system/libsystem_c.dylib)
==48994==    by 0x17F2CF: printf (in /usr/lib/system/libsystem_c.dylib)
==48994==    by 0x100001AFE: main (bst.c:78)
==48994== 
BST Dump: 
Tree: 0x7FFF5FBFF410 (root 0x000000000, height 0)
hi1
BST Dump: After  0: value 19
Tree: 0x7FFF5FBFF410 (root 0x100009170, height 0)
Number:   19 - Node: 0x100009170 (parent 0x000000000, left 0x000000000, right 0x000000000)
hi2
BST Dump: After  1: value  7
Tree: 0x7FFF5FBFF410 (root 0x100009170, height 0)
Number:   19 - Node: 0x100009170 (parent 0x000000000, left 0x1000091D0, right 0x000000000)
Number:    7 - Node: 0x1000091D0 (parent 0x100009170, left 0x000000000, right 0x000000000)
hi2
hi3
BST Dump: After  2: value 12
Tree: 0x7FFF5FBFF410 (root 0x100009170, height 0)
Number:   19 - Node: 0x100009170 (parent 0x000000000, left 0x1000091D0, right 0x000000000)
Number:    7 - Node: 0x1000091D0 (parent 0x100009170, left 0x000000000, right 0x100009230)
Number:   12 - Node: 0x100009230 (parent 0x1000091D0, left 0x000000000, right 0x000000000)
hi2
hi3
hi3
BST Dump: After  3: value 15
Tree: 0x7FFF5FBFF410 (root 0x100009170, height 0)
Number:   19 - Node: 0x100009170 (parent 0x000000000, left 0x1000091D0, right 0x000000000)
Number:    7 - Node: 0x1000091D0 (parent 0x100009170, left 0x000000000, right 0x100009230)
Number:   12 - Node: 0x100009230 (parent 0x1000091D0, left 0x000000000, right 0x100009290)
Number:   15 - Node: 0x100009290 (parent 0x100009230, left 0x000000000, right 0x000000000)
hi3
BST Dump: After  4: value 21
Tree: 0x7FFF5FBFF410 (root 0x100009170, height 0)
Number:   19 - Node: 0x100009170 (parent 0x000000000, left 0x1000091D0, right 0x1000092F0)
Number:    7 - Node: 0x1000091D0 (parent 0x100009170, left 0x000000000, right 0x100009230)
Number:   12 - Node: 0x100009230 (parent 0x1000091D0, left 0x000000000, right 0x100009290)
Number:   15 - Node: 0x100009290 (parent 0x100009230, left 0x000000000, right 0x000000000)
Number:   21 - Node: 0x1000092F0 (parent 0x100009170, left 0x000000000, right 0x000000000)
hi2
hi3
hi2
BST Dump: After  5: value  8
Tree: 0x7FFF5FBFF410 (root 0x100009170, height 0)
Number:   19 - Node: 0x100009170 (parent 0x000000000, left 0x1000091D0, right 0x1000092F0)
Number:    7 - Node: 0x1000091D0 (parent 0x100009170, left 0x000000000, right 0x100009230)
Number:   12 - Node: 0x100009230 (parent 0x1000091D0, left 0x100009350, right 0x100009290)
Number:    8 - Node: 0x100009350 (parent 0x100009230, left 0x000000000, right 0x000000000)
Number:   15 - Node: 0x100009290 (parent 0x100009230, left 0x000000000, right 0x000000000)
Number:   21 - Node: 0x1000092F0 (parent 0x100009170, left 0x000000000, right 0x000000000)
hi2
hi3
hi3
hi3
BST Dump: After  6: value 16
Tree: 0x7FFF5FBFF410 (root 0x100009170, height 0)
Number:   19 - Node: 0x100009170 (parent 0x000000000, left 0x1000091D0, right 0x1000092F0)
Number:    7 - Node: 0x1000091D0 (parent 0x100009170, left 0x000000000, right 0x100009230)
Number:   12 - Node: 0x100009230 (parent 0x1000091D0, left 0x100009350, right 0x100009290)
Number:    8 - Node: 0x100009350 (parent 0x100009230, left 0x000000000, right 0x000000000)
Number:   15 - Node: 0x100009290 (parent 0x100009230, left 0x000000000, right 0x1000093B0)
Number:   16 - Node: 0x1000093B0 (parent 0x100009290, left 0x000000000, right 0x000000000)
Number:   21 - Node: 0x1000092F0 (parent 0x100009170, left 0x000000000, right 0x000000000)
hi2
hi2
BST Dump: After  7: value  1
Tree: 0x7FFF5FBFF410 (root 0x100009170, height 0)
Number:   19 - Node: 0x100009170 (parent 0x000000000, left 0x1000091D0, right 0x1000092F0)
Number:    7 - Node: 0x1000091D0 (parent 0x100009170, left 0x100009410, right 0x100009230)
Number:    1 - Node: 0x100009410 (parent 0x1000091D0, left 0x000000000, right 0x000000000)
Number:   12 - Node: 0x100009230 (parent 0x1000091D0, left 0x100009350, right 0x100009290)
Number:    8 - Node: 0x100009350 (parent 0x100009230, left 0x000000000, right 0x000000000)
Number:   15 - Node: 0x100009290 (parent 0x100009230, left 0x000000000, right 0x1000093B0)
Number:   16 - Node: 0x1000093B0 (parent 0x100009290, left 0x000000000, right 0x000000000)
Number:   21 - Node: 0x1000092F0 (parent 0x100009170, left 0x000000000, right 0x000000000)
hi2
hi2
hi3
BST Dump: After  8: value  3
Tree: 0x7FFF5FBFF410 (root 0x100009170, height 0)
Number:   19 - Node: 0x100009170 (parent 0x000000000, left 0x1000091D0, right 0x1000092F0)
Number:    7 - Node: 0x1000091D0 (parent 0x100009170, left 0x100009410, right 0x100009230)
Number:    1 - Node: 0x100009410 (parent 0x1000091D0, left 0x000000000, right 0x100009470)
Number:    3 - Node: 0x100009470 (parent 0x100009410, left 0x000000000, right 0x000000000)
Number:   12 - Node: 0x100009230 (parent 0x1000091D0, left 0x100009350, right 0x100009290)
Number:    8 - Node: 0x100009350 (parent 0x100009230, left 0x000000000, right 0x000000000)
Number:   15 - Node: 0x100009290 (parent 0x100009230, left 0x000000000, right 0x1000093B0)
Number:   16 - Node: 0x1000093B0 (parent 0x100009290, left 0x000000000, right 0x000000000)
Number:   21 - Node: 0x1000092F0 (parent 0x100009170, left 0x000000000, right 0x000000000)
hi2
hi2
hi3
hi3
BST Dump: After  9: value  5
Tree: 0x7FFF5FBFF410 (root 0x100009170, height 0)
Number:   19 - Node: 0x100009170 (parent 0x000000000, left 0x1000091D0, right 0x1000092F0)
Number:    7 - Node: 0x1000091D0 (parent 0x100009170, left 0x100009410, right 0x100009230)
Number:    1 - Node: 0x100009410 (parent 0x1000091D0, left 0x000000000, right 0x100009470)
Number:    3 - Node: 0x100009470 (parent 0x100009410, left 0x000000000, right 0x1000094D0)
Number:    5 - Node: 0x1000094D0 (parent 0x100009470, left 0x000000000, right 0x000000000)
Number:   12 - Node: 0x100009230 (parent 0x1000091D0, left 0x100009350, right 0x100009290)
Number:    8 - Node: 0x100009350 (parent 0x100009230, left 0x000000000, right 0x000000000)
Number:   15 - Node: 0x100009290 (parent 0x100009230, left 0x000000000, right 0x1000093B0)
Number:   16 - Node: 0x1000093B0 (parent 0x100009290, left 0x000000000, right 0x000000000)
Number:   21 - Node: 0x1000092F0 (parent 0x100009170, left 0x000000000, right 0x000000000)
==48994== 
==48994== HEAP SUMMARY:
==48994==     in use at exit: 18,500 bytes in 33 blocks
==48994==   total heap usage: 43 allocs, 10 frees, 18,820 bytes allocated
==48994== 
==48994== LEAK SUMMARY:
==48994==    definitely lost: 0 bytes in 0 blocks
==48994==    indirectly lost: 0 bytes in 0 blocks
==48994==      possibly lost: 0 bytes in 0 blocks
==48994==    still reachable: 18,500 bytes in 33 blocks
==48994==         suppressed: 0 bytes in 0 blocks
==48994== Rerun with --leak-check=full to see details of leaked memory
==48994== 
==48994== For counts of detected and suppressed errors, rerun with: -v
==48994== Use --track-origins=yes to see where uninitialised values come from
==48994== ERROR SUMMARY: 1 errors from 1 contexts (suppressed: 0 from 0)

这台机器上仍在使用的 33 个分配是完全正常的（Mac OS X 10.7.5，Valgrind 3.7.1）。它们出现在调用之前调用的函数main()中。

除了 中的系统更改Compare()和动态分配之外AddNode()，代码还不错。

于 2013-04-04T02:02:16.957 回答
0

我可以看到的问题之一是您的代码

void AddNode(int num, Tree * tree){
    Node NewNode;
    NewNode.Num=num;
    NewNode.left=NULL;
    NewNode.right=NULL;
    NewNode.parent=NULL;

    Compare(tree,&NewNode,tree->root);
}

您将 newNode 声明为简单的局部变量，因此一旦您退出 AddNode 函数（在调用 compare 之后），变量 newNode 就会超出范围并且它的值丢失。

您可以尝试使用指针，使用 malloc 分配空间，然后添加新节点

于 2013-04-04T02:04:57.290 回答