Question

我正在编写程序的一部分，在该程序中遍历二叉树，直到在树中找到想要的项目为止（假设我们要查找的项目始终存在于树中）。我在树中搜索的方法是预定的树遍历。

每个节点包含一个问题或一个陈述。如果节点为statement，则它没有子节点。但是，如果该节点是一个问题，则它恰好有两个孩子。在下面的内容中，您可以看到用于创建节点的头文件。

// two type of branch in the tree
enum response{
    YES,
    NO
};

// content of a node
union objInfoOrQInfo {
    char * object;
    char * question;
};

struct node{
    union objInfoOrQInfo container;
   // enum response existLeftChild;
   // enum response existRightChild;
    struct node * rightChild;
    struct node * leftChild;
};

void nodePrint(char * string);

在下面的代码中，我填充了一棵树并实现了预定的树遍历

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include "node.h"

struct node * traverse(struct node * searchingNode, char * searchContent);


void nodePrint(char * string)
{
    struct node root;
    struct node left;
    struct node right;
    //second level
    struct node leftLeft;
    struct node leftRight;
    struct node rightLeft;
    struct node rightRight;

    // Populating the tree
    root.container.question = "Does it have a tail?";
    root.leftChild = &left;
    root.rightChild = &right;

    left.container.question = "Does it like a chase mice?";
    left.leftChild = &leftLeft;
    left.leftChild = &leftRight;
    right.container.question = "Is it flat, round and edible?";
    right.leftChild = &rightLeft;
    right.rightChild = &rightRight;

    leftLeft.container.object = "A Cat";
    leftLeft.leftChild = NULL;
    leftLeft.rightChild = NULL;
    leftRight.container.object = "A Pangolin";
    leftRight.leftChild = NULL;
    leftRight.rightChild = NULL;
    rightLeft.container.object = "A Pizza";
    rightLeft.leftChild = NULL;
    rightLeft.rightChild = NULL;
    rightRight.container.object = "Pete";
    rightLeft.leftChild = NULL;
    rightLeft.rightChild = NULL;


    // It   needs to traverse from the root until it finds the content you are asking for
    struct node * result = traverse(&root, string);
    if(result->leftChild != NULL || result->rightChild != NULL)
    {
        printf("Object: %s\n","[NOTHING]");
        printf("Question: %s\n",result->container.question);
        if(result->leftChild->container.object == NULL)
            printf("Yes: %s\n",result->leftChild->container.question);
        else
            printf("Yes: %s\n",result->leftChild->container.object);

        if(result->leftChild->container.object == NULL)
            printf("Yes: %s\n",result->rightChild->container.question);
        else
            printf("Yes: %s\n",result->rightChild->container.object);
    }
    else
    {
        printf("Object: %s\n","[NOTHING]");
        printf("Question: %s\n",result->container.question);
    }


}

// This function traverse from the root until it finds the desired content (return 1), otherwise (return -1)
struct node * traverse(struct node * searchingNode, char * searchContent) {

    // If the node is found, just return it.
    if (strcmp(searchingNode->container.question, searchContent) == 0)
        return searchingNode;

    // If the node is not found and we are at a leaf, return NIL
    if (searchingNode->leftChild == NULL && searchingNode->rightChild == NULL)
        return NULL;

    // Searching in the left and right subtrees, respectively.
    struct node *leftSubTree = traverse(searchingNode->leftChild, searchContent);
    struct node *rightSubTree;
    if(leftSubTree == NULL)
        rightSubTree = traverse(searchingNode->rightChild, searchContent);

    // Calculating the final result
    if (leftSubTree == NULL && rightSubTree == NULL)
        return NULL;
    else if (leftSubTree == NULL)
        return rightSubTree;
    return leftSubTree;
}

为了搜索节点，我需要调用nodePrint（）并在主函数中将适当的参数传递给它，如下所示：

int main() {

    nodePrint("A Pizza");

    return 0;
}

如果节点数据是一个问题，则预期结果应为：

Object: [NOTHING]
Question: Is it flat, round and edible?
Yes: A Pizza
Yes: Pete

如果数据是语句，则输出应为：

Object: [NOTHING]
Question: A Pizza

我的程序显示了适当的结果，但是有一个SEGFAULT，我无法弄清楚是什么原因引起的。我真的很感谢有人能帮我调试它。

二叉搜索树实现遍历时的分割错误

0 个答案: