有效地打印树节点及其所有子节点

Print tree node and all of it's childs efficiently

我正在尝试创建一个函数来打印一个节点及其所有子节点,但我正在尝试使其高效且递归。但实际上并没有用。

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

#define SIZE    100

typedef struct tree {
    int value;
    struct tree *child, *sibling, *parent;
} *Tree;

Tree initTree(int value) {
    Tree root = malloc(sizeof(struct tree));
    root->value = value;
    root->parent = NULL;
    root->child = NULL;
    root->sibling = NULL;
    return root;
}

void drawTreeHelper(Tree tree, FILE* stream) {
    Tree tmp;
    if (tree == NULL) {
        return;
    }
    fprintf(stream, "    %ld[label=\"%d\", fillcolor=red]\n", (intptr_t) tree, tree->value);
    tmp = tree->child;

    while (tmp != NULL) {
        fprintf(stream, "    %ld -> %ld \n", (intptr_t) tree, (intptr_t) tmp);
        drawTreeHelper(tmp, stream);
        tmp = tmp->sibling;
    }
}

void drawTree(Tree tree, char *fileName) {
    FILE* stream = fopen("test.dot", "w");
    char buffer[SIZE];
    fprintf(stream, "digraph tree {\n");
    fprintf(stream, "    node [fontname=\"Arial\", shape=circle, style=filled, fillcolor=yellow];\n");
    if (tree == NULL)
        fprintf(stream, "\n");
    else if (!tree->child)
        fprintf(stream, "    %ld [label=\"%d\"];\n", (intptr_t) tree, tree->value);
    else
        drawTreeHelper(tree, stream);
    fprintf(stream, "}\n");
    fclose(stream);
    sprintf(buffer, "dot test.dot | neato -n -Tpng -o %s", fileName);
    system(buffer);
}

Tree uniteTries(Tree child, Tree parent)
{
    if (parent)
    {
        if (!parent->child) parent->child = child;
        else
        {
            Tree iter = parent->child;
            while (iter->sibling) iter = iter->sibling;
            iter->sibling = child;
        }
    }
    return parent;
}

Tree uniteForest(Tree root, Tree *forest, int n)
{
    int i;
    for (i = 0; i < n; ++i)
    {
        if (forest[i]) root = uniteTries(forest[i], forest[i]->parent);
    }
    root = forest[0];
    return root;
}

void printParentChildRec(Tree root)
{
    if(!root) return;
    printf("%d ", root->value);

    printParentChildRec(root->sibling);
    printParentChildRec(root->child);
}

int main() {
    int i;
    char buffer[SIZE];
    Tree *forest = malloc(6 * sizeof(Tree));
    for (i = 0; i < 6; i++) {
        forest[i] = initTree(i);
    }

    forest[1]->parent = forest[0];
    forest[2]->parent = forest[0];
    forest[3]->parent = forest[0];
    forest[4]->parent = forest[1];
    forest[5]->parent = forest[1];

    Tree root = uniteForest(root, forest, 6);

    printParentChildRec(root);

    drawTree(root, "tree.png");

    return 0;
}

此代码将为您提供一个可验证的示例,这是我尝试做的事情:

void printParentChildRec(Tree root) {
    if (!root)
        return;
    printf("%d ", root->value);

    printParentChildRec(root->sibling);
    printParentChildRec(root->child);
}

我得到的结果只是 0 1 2 3 4 5,这是所有节点,但我想打印如下内容:

0 1 2 3
1 4 5
2
3
4
5

您的代码中存在一些问题:

  • 您将指针隐藏在 typedef 后面,这会让代码的读者感到困惑,并且经常会导致编程错误。
  • 您使用 %ld 打印 intptr_t 值,这在 intptr_t 不是 long 的别名并且具有不同大小的平台上是未定义的行为,例如Windows 64 位。此类型没有特定的 printf 格式,您应该将值重铸为 (long)(intptr_t)tree(long long)(intptr_t)tree 并使用 %lld 或其未签名版本或使用 %p 格式为 (void *)tree.
  • 您期望的结果不是以文本形式生成,而是某种形式的图形渲染,很难从代码中分析出来。首先生成文本输出会更容易调试。

您的代码中还有更多问题:

  • main()中,Tree root = uniteForest(root, forest, 6);将未定义的变量root传递给uniteForest
  • argument root in Tree uniteForest(Tree root, Tree *forest, int n) 从未使用过,它仅用于存储临时结果。您应该只删除参数并将代码简化为:

    Tree uniteForest(Tree *forest, int n) {
        for (int i = 0; i < n; i++) {
            if (forest[i])
                uniteTries(forest[i], forest[i]->parent);
        }
        return forest[0];
    }
    
  • main 只打印树的根,因此递归地打印 forest[0] 及其后代的值。相反,您想要打印节点的值及其直接子节点的值,然后 为每个子节点递归。

这是更正后的版本:

void printParentChildRec(Tree node) {
    if (node) {
        printf("%d ", node->value);
        for (Tree child = node->child; child; child = child->sibling) {
            printf("%d ", child->value);
        }
        printf("\n");
        for (Tree child = node->child; child; child = child->sibling) {
            printParentChildRec(child);
        }
    }
}

输出:

0 1 2 3
1 4 5
4
5
2
3

我认为您的统一树功能是这里的问题。我 运行 使用我的调试器编写了这段代码,这就是它从 root

开始的样子

这是联合树方法后输入打印函数的根:

你能告诉我你想在这里实现什么,也许我可以帮助你!