Question

我尝试编写以下功能

typedef struct node{
    char *name;
    int n;  //number of kids
    struct node **kids;
    int dtype; // COMPOSITE or BASIC
    union data{
        double price; //for BASIC
        char *quantity; //for COMPOSITE
    }data;
}node;

void create_kid(node **parent){
    if(*parent == NULL){
        (*parent) = malloc(sizeof(node));
        (*parent)->n = 0;
        (*parent)->kids = NULL;
    }else{
        (*parent)->n += 1;
        (*parent)->kids = realloc((*parent)->kids, ((*parent)->n)*(sizeof(node)));
        (*parent)->kids[(*parent)->n - 1] = malloc(sizeof(node));
        (*parent)->kids[(*parent)->n - 1]->n = 0;
        (*parent)->kids[(*parent)->n - 1]->kids = NULL;
    }
}

void insert_c(node *node, char *str){
        node->name = malloc(MAX);
        node->name = str;
        node->dtype = COMPOSITE;
}

void insert_b(node *node, char *str){
        node->name = malloc(MAX);
        node->name = str;
        node->dtype = BASIC;
}

用这样的输入来制作这样的n-ary树

BIKE(2*WHEEL(RIM[60.0 ],
        2*AXLE,
        SPOKE[120.],
        HUB(2*GEAR[25.],AXLE(5*BOLT[0.1], 7 * NUT[.15]))),
    FRAME(REARFRAME [175.00],
        1*FRONTFRAME (FORK[22.5] ,AXLE, 2 *HANDLE[10.])))

我知道n-ary树的其他表示，其中包括*兄弟姐妹和* firstkid，但我相信这种**孩子表示更适合我的情况。我想问一些事情。

首先，这些功能有什么问题吗？它们是否适合构建一棵n-ary树？

其次，即使我的功能正确，我也无法从输入构造树。例如，根据输入，如果我一次只读一个单词，函数调用必须与我的函数一样：

create_kid(&tree);
insert_c(tree, "BIKE");
create_kid(&tree);
insert_c(tree->kids[0], "WHEEL");
create_kid(&(tree->kids[0]));
insert_c(tree->kids[0]->kids[0], "RIM");
.
.
.
create_kid(&tree);
insert_c(tree->kids[1], "FRAME");
.
.

如果以这种方式形成树，我需要从更高的深度到更低的深度，例如从NUT到FRAME。所以我认为必须有一种更简单的方法，也许是递归方法。我有什么方法可以通过递归来做到这一点吗？

Answer 1

首先是小事：

如果您的AXLE节点的相同实例位于“树”中的不同位置，则您的n-ary树不再是n-ary树。这是一张图。如果您的Axle，还包含一辆自行车，它甚至可以是一个循环图。

实施所有这些时你会发现不同之处。您将需要引用计数或图表中的某些内容来决定何时真正释放节点。在真正的n-ary树中，您不需要它，因为每个节点只在树中出现一次。

下一步管理事项：

您决定使用指针数组作为数据结构来包含节点的子节点（孩子）。因此，为了降低代码的复杂性，实际创建（动态）指针数组作为第一步是个好主意。好处：您的树代码不会受到动态数组代码的负担，您可以进行单元测试，甚至可以为其他项目重复使用动态指针数组。

我敲了一个示例指针数组实现（没有经过深度测试）以显示原理。

为了让这里显示的代码更加“用户友好 - 首先是我使用的包含，为了解决这个问题：

#include <crtdbg.h> // Windows specific - helps find memory leaks.
#include <stdint.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>

现在到指针数组实现：

typedef struct PointerArray_tag
{
    void**data;
    size_t capacity;
    size_t size;
} PointerArray_t;

void PointerArrayInitialize(PointerArray_t *array)
{
    array->data = NULL;
    array->capacity = 0;
    array->size = 0;
}

int PointerArrayReserve(PointerArray_t *array, size_t capacity)
{
    if (array->capacity < capacity)
    {
        void **newData = (void**)realloc(array->data, capacity * sizeof(void*));
        if (NULL != newData)
        {
            array->data = newData;
            array->capacity = capacity;
            return 1; // 1 = "true" in C and indicates success.
        }
        return 0; // 0 = "false" in C and indicates failure...
    }
    else
    {
        return 1;
    }
}
typedef void (*PointerArrayElementFree_t)(void *element);

int PointerArrayResize(PointerArray_t *array, size_t newSize, void* value, PointerArrayElementFree_t elementFree)
{
    if (NULL == array) return 0;
    if (NULL == elementFree) return 0;

    if (newSize < array->size)
    {
        for (size_t index = newSize; index < array->size; index++)
        {
            elementFree(array->data[index]);
        }
        array->size = newSize;
        return 1;
    }
    else if (newSize > array->size)
    {
        size_t oldSize = array->size;
        if (PointerArrayReserve(array, newSize))
        {
            for (size_t index = oldSize; index < newSize; index++)
            {
                array->data[index] = value;
            }
            array->size = newSize;
            return 1;
        }
        return 0; // Reserve() failed, so this function also failed.
    }
    else
        return 1; // oldSize == newSize - nothing to do.
}

int PointerArrayInitializeWithCapacity(PointerArray_t *array, size_t initialCapacity)
{
    PointerArrayInitialize(array);
    return PointerArrayReserve(array, initialCapacity);
}

int PointerArrayPushBack(PointerArray_t *array, void*element)
{
    if (PointerArrayReserve(array, array->size + 1))
    {
        array->data[array->size] = element;
        array->size++;
        return 1;
    }
    return 0;
}

int PointerArrayClear(PointerArray_t *array, PointerArrayElementFree_t elementFree)
{
    if (NULL == elementFree) return 0;
    if (NULL == array) return 0;

    for (size_t index = 0; index < array->size; index++)
    {
        elementFree(array->data[index]);
        array->data[index] = NULL;
    }
    array->size = 0;
    return 1;
}

int PointerArrayUninitialize(PointerArray_t *array, PointerArrayElementFree_t elementFree)
{
    if (PointerArrayClear(array, elementFree))
    {
        free(array->data);
        array->data = NULL;
        array->capacity = 0;
        return 1;
    }
    return 0;
}

这里没什么特别的。具有常规操作的void指针的动态数组。

接下来，虽然不是100％适合你的问题（我觉得这将花费我更多的时间去写信），n-ary树实现。如上所述，还没有准备好作为图表。

此外，我冒昧地采取了一种语法形式：

<Material> ::= <Count> <Name> <Size>
            |  <Count> <Name>

解析此产品，解析器首先会找到一个数字（例如2个轮子），然后找到材料的名称（轮子），然后可选择测量（例如Nuts [0.15]）。

当应用程序使用AST并且过于复杂时，更接近你的问题的替代语法将更难处理：

<Material> ::= <Count> <Name> <Size>
            | <Count> <Name>
            | <Name>
            | <Name> <Size>

这暗示如果你对输入文本的语法有影响，你可以改进它。

所以，这里是ComponentTree代码，它具有上述约束。

enum NodeContentType
{
    EMPTY = 0
,   NUMBER = 1
,   NAME = 2
};

typedef struct NodeContent_tag
{
    NodeContentType type;
    union
    {
        float number;
        char *name;
    };
} NodeContent_t;

typedef struct ComponentNode_tag
{
    NodeContent_t content;
    ComponentNode_tag * parent;
    PointerArray_t children;
} ComponentNode_t;

void ComponentNodeFree(ComponentNode_t *node)
{
    // Clean up anything heap based in NodeContent_t.
    switch (node->content.type)
    {
    case NAME:
        free(node->content.name);
        node->content.name = NULL;
        node->content.type = EMPTY;
        break;
    default:
        // nothing to do.
        break;
    }

    // Free all the children below this node recursively.
    PointerArrayUninitialize(&node->children, (PointerArrayElementFree_t)ComponentNodeFree);
    free(node);
}
typedef struct ComponentTree_tag
{
    ComponentNode_t root;
} ComponentTree_t;

void ComponentTreeInitialize(ComponentTree_t *tree)
{
    tree->root.content.type = EMPTY;
    PointerArrayInitialize(&tree->root.children);
    tree->root.parent = NULL;
}

ComponentNode_t *ComponentTreeRoot(ComponentTree_t *tree)
{
    return &tree->root;
}

void ComponentTreeClear(ComponentTree_t *tree)
{
    if (NULL != tree)
    {
        PointerArrayClear(&tree->root.children,(PointerArrayElementFree_t)ComponentNodeFree);
    }
}

void ComponentTreeUninitialize(ComponentTree_t *tree)
{
    if (NULL != tree)
    {
        PointerArrayUninitialize(&tree->root.children, (PointerArrayElementFree_t)ComponentNodeFree);
    }
}

ComponentNode_t *ComponentTreeCreateEmptyNode()
{
    ComponentNode_t *node = (ComponentNode_t*)malloc(sizeof(ComponentNode_t));
    if (NULL != node)
    {
        node->content.type = EMPTY;
        node->parent = NULL;
        PointerArrayInitialize(&node->children);
    }
    return node;
}

ComponentNode_t *ComponentTreeCreateNameNode(const char *name)
{
    ComponentNode_t *node = (ComponentNode_t*)malloc(sizeof(ComponentNode_t));
    if (NULL != node)
    {
        node->content.type = NAME;
        node->content.name = _strdup(name);
        assert(NULL != node->content.name); 
        if (NULL == node->content.name)
        {
            free(node);
            return NULL;
        }
        node->parent = NULL;
        PointerArrayInitialize(&node->children);
    }
    return node;
}

ComponentNode_t *ComponentTreeCreateNumberNode(float number)
{
    ComponentNode_t *node = (ComponentNode_t*)malloc(sizeof(ComponentNode_t));
    if (NULL != node)
    {
        node->content.type = NUMBER;
        node->content.number = number;
        node->parent = NULL;
        PointerArrayInitialize(&node->children);
    }
    return node;
}


int ComponentTreeJoin(ComponentTree_t *tree, ComponentNode_t *where, ComponentNode_t *child)
{
    if (NULL == child) return 0;
    if (NULL != child->parent) return 0; // is already in a tree.
    if (NULL == tree) return 0;
    if (NULL == where)
    {
        where = &tree->root;
    }
    child->parent = where;
    return PointerArrayPushBack(&where->children, child);
}

乍一看，ComponentTree_t *tree中存在ComponentTreeJoin()参数可能看起来有点滑稽。相反，CompoenentTreeCreateXXXNode()函数也应该具有它。为什么？因为以后有人可能希望拥有ComponentTree_t *ComponentTreeFromNode(ComponentNode_t *node)或ComponentNode_t *ComponentTreeGetRoot(ComponentNode_t *node)这样的功能。

最后，最重要的是，树的手动装配看起来像这样（不完整，这个答案已经很长）：

int _tmain(int argc, _TCHAR* argv[])
{
    ComponentTree_t bikeTree;
    ComponentTreeInitialize(&bikeTree);
    ComponentNode_t * nutSize = ComponentTreeCreateNumberNode(0.15f);
    ComponentNode_t * nut = ComponentTreeCreateNameNode("NUT");
    ComponentTreeJoin(&bikeTree, nut, nutSize);
    ComponentNode_t *nutCount = ComponentTreeCreateNumberNode(7.0f);
    ComponentTreeJoin(&bikeTree, nutCount, nut);
    ComponentNode_t * boltSize = ComponentTreeCreateNumberNode(0.1f);
    ComponentNode_t * bolt = ComponentTreeCreateNameNode("BOLT");
    ComponentTreeJoin(&bikeTree, bolt, boltSize);
    ComponentNode_t * boltCount = ComponentTreeCreateNumberNode(5.0f);
    ComponentTreeJoin(&bikeTree, boltCount, bolt);
    ComponentNode_t * axle = ComponentTreeCreateNameNode("AXLE");
    ComponentTreeJoin(&bikeTree, axle, nutCount);
    ComponentTreeJoin(&bikeTree, axle, boltCount);
    ComponentNode_t *axleCount = ComponentTreeCreateNumberNode(1.0f);
    ComponentTreeJoin(&bikeTree, axleCount, axle);
    ComponentNode_t *gearSize = ComponentTreeCreateNumberNode(25.0f);
    ComponentNode_t *gear = ComponentTreeCreateNameNode("GEAR");
    ComponentTreeJoin(&bikeTree, gear, gearSize);
    ComponentNode_t *gearCount = ComponentTreeCreateNumberNode(2.0f);
    ComponentTreeJoin(&bikeTree, gearCount, gear);
    ComponentNode_t *hub = ComponentTreeCreateNameNode("HUB");
    ComponentTreeJoin(&bikeTree, hub, gearCount);
    ComponentTreeJoin(&bikeTree, hub, axleCount);
    ComponentNode_t *hubCount = ComponentTreeCreateNumberNode(1.0f);
    ComponentTreeJoin(&bikeTree, hubCount, hub);
    ComponentNode_t * spokeSize = ComponentTreeCreateNumberNode(120.0f);
    ComponentNode_t * spoke = ComponentTreeCreateNameNode("SPOKE");
    ComponentTreeJoin(&bikeTree, spoke, spokeSize);
    ComponentNode_t * spokeCount = ComponentTreeCreateNumberNode(1.0f);
    ComponentTreeJoin(&bikeTree, spokeCount, spoke);
    ComponentNode_t * axle1 = ComponentTreeCreateNameNode("AXLE");
    ComponentNode_t * axle1Count = ComponentTreeCreateNumberNode(2.0f);
    ComponentTreeJoin(&bikeTree, axle1Count, axle1);
    // ...
    ComponentNode_t * wheel = ComponentTreeCreateNameNode("WHEEL");
    ComponentTreeJoin(&bikeTree, wheel, axle1Count);
    ComponentTreeJoin(&bikeTree, wheel, spokeCount);
    ComponentTreeJoin(&bikeTree, wheel, hubCount);
    ComponentNode_t * wheelCount = ComponentTreeCreateNumberNode(2.0f);
    ComponentTreeJoin(&bikeTree, wheelCount, wheel);
    ComponentNode_t *bike = ComponentTreeCreateNameNode("BIKE");
    ComponentTreeJoin(&bikeTree, bike, wheelCount);
    ComponentNode_t *bikeCount = ComponentTreeCreateNumberNode(1.0f);
    ComponentTreeJoin(&bikeTree, bikeCount, bike);
    // .. frame branch omitted .. 
    ComponentTreeJoin(&bikeTree, NULL, bikeCount);
    ComponentTreeUninitialize(&bikeTree);

    _CrtDumpMemoryLeaks();
    return 0;
}

要点：

如果真的应该是图表，而不是n-ary树，那么有人选择了错误的术语。
混合到树实现中的动态数组代码会降低代码的可测试性和可读性。
如果使用动态数组指针作为kids-list，或者单个链接列表不是真正的基础。可能列表更适合，因为示例输入中的大多数节点无论如何只有1个子节点，并且差异是边际的。数组的当前实现也没有节省堆操作量（但如果默认容量没有被编程为0但可能是1或2或其他适当的值，它可能会。）
main()代码显示了如何构造树的方式与解析器遍历输入文本的顺序不完全相同。（我有时从右到左而不是100％从左到右）。
此处显示的代码不会将物料项的数量视为节点的成员，但会在物料名称节点的顶部生成一个数字节点。这可以但不一定是有益的，因为它完全取决于语法的定义方式和解析器的类型等。通常情况就是如此。 AST最终看起来如何也是由解析器决策驱动的。

N-ary树使用stdin函数（makenode，insert）

1 个答案: