我的二叉树程序在输出时崩溃

2022-02-23 00:00:00 tree binary-search-tree binary-tree c++

我正在尝试编写二叉树,但它崩溃了。 它一直打印出相同的数字:左节点(较小的数字)和根节点。它递归地重复该程序,直到崩溃。 我知道我的代码还有其他一些错误,但这是需要解决的主要问题,所以我的代码如下:

#include <iostream>

using namespace std;

struct node {
    int value = 0;
    node* left = NULL;
    node* right = NULL;
};

node root;
void add(int x, node* curr)
{

    if (x < (*curr).value) {

        if ((*curr).left == NULL) {
            node next;
            next.value = x;
            (*curr).left = &next;
        }
        else {
            add(x, (*curr).left);
        }

        if (x > (*curr).value) {
            if ((*curr).right == NULL) {
                node next;
                next.value = x;
                (*curr).right = &next;
            }
            else {
                add(x, (*curr).right);
            }
        }
    }
}

void out(node ro)
{
    node lefta;
    node righta;
    if (ro.left != NULL) {
        lefta = *(ro.left);
        cout << lefta.value;
        out(lefta);
    }
    if (ro.right != NULL) {
        righta = *(ro.right);
        cout << " " << righta.value << endl;
        out(righta);
    }
}

int main()
{

    int n;
    cin >> n;
    int x;
    cin >> x;
    root.value = x;
    node* curr;
    curr = &root;
    for (int i = 1; i < n; i++) {
        cin >> x;
        add(x, curr);
    }

    out(root);

    return 0;
}

崩溃后返回:

Process returned -1073741571 (0xC00000FD)

解决方案

如果您想要将新节点插入到列表中,则必须对其进行分配。 函数返回时,堆栈上的局部变量超出范围。 无论如何,您必须将您的函数add调整为适用于如下所示的双向链表:

void add(int x, node *curr)
{
    // while x less than curr->value step left 
    while ( curr->left != NULL && x < curr->value )
        curr = curr->left;

    // while x greater than curr->next->value step right 
    while ( curr->right != NULL && x > curr->right->value )
        curr = curr->right;

    // x is less than curr->right->value (curr->right may be NULL)
    // either x is greater curr->value or curr->left == NULL 

    node *newNode = new node; // allcat new node
    newNode->left = newNode->right = NULL; // <- this schould be done by a constructor of node
    newNode->value = x;

    if ( x < curr->value )
    {
        // curr->left == NULL => new node is new start of list
        curr->left = newNode;
        newNode->right = curr;
    }
    else if ( curr->right == NULL )
    {
        // new node is new end of list
        curr->right = newNode;
        newNode->left = curr;
    }
    else
    {
        // new node someweher in the list
        node *rightNode = curr->right;
        curr->right = newNode;
        newNode->right = rightNode;
        newNode->left = curr;
        rightNode->left = newNode;
    }
}

注意,您使用new分配的所有节点都必须delete销毁。

如果您希望从头到尾打印列表,则不需要递归函数,并且避免复制节点。使用指针:

void out(const node *ro)
{
    if ( ro == NULL )
        return;

    while ( ro->left != NULL )
        ro = ro->left;

    while ( ro != NULL )
    {
        cout << ro->value;
        ro = ro->right;
    }
}

...

out(&root);

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