二叉树非递归遍历内存不对齐

问题遇到的现象和发生背景

leetcode144 非递归遍历二叉数
发生member access within misaligned address错误

用代码块功能插入代码，请勿粘贴截图

struct stack {
    struct TreeNode* root;
    struct stack* next;
};

//在cur后面添加栈元素
void push_back(struct TreeNode* root, struct stack* cur) {
    struct stack* nextNode = (struct stack*)malloc(sizeof(struct stack));
    nextNode->root = root;
    nextNode->next = NULL;
    cur->next = nextNode;
    cur = cur->next;
}
void popTop(struct stack* shead) {
    struct stack* slow = shead;
    struct stack* fast = slow->next;
    if (!fast) return;
    while (fast->next)
    {
        slow = slow->next;
        fast = fast->next;
    }
    free(fast);
    slow->next = NULL;
}


int* preorderTraversal(struct TreeNode* root, int* returnSize) {
    int ret[1000];
    int retIndex = 0;
    struct stack* shead = (struct stack*)malloc(sizeof(struct stack));
    shead->next = NULL;
    struct stack* cur = shead;
    int stackTop = 0;
    *returnSize = 0;

    if (!root) return ret;

    push_back(root, cur);

    //当栈不为空时
    while (shead->next)
    {
        struct TreeNode* node = cur->root;
        ret[retIndex++] = node->val;
        popTop(shead);
        if (node->right) push_back(node->right, cur);
        if (node->left) push_back(node->left, cur);
    }

    *returnSize = retIndex;
    return ret;
}

运行结果及报错内容

实在是找不到问题出在哪里

TreeNode的定义呢

• 这篇文章：不使用递归遍历二叉树 也许能够解决你的问题，你可以看下

解决了，是用链表模拟栈的相关代码有问题
修改之后的代码

struct stack {
    struct TreeNode* root;
    struct stack* next;
};
struct TreeNode {
    int val;
    struct TreeNode* left;
    struct TreeNode* right;
};
//在栈顶添加元素
void push_back(struct TreeNode* root, struct stack* shead) {
    struct stack* cur = shead;
    while (cur->next)
        cur = cur->next;

    struct stack* nextNode = (struct stack*)malloc(sizeof(struct stack));
    nextNode->root = root;
    nextNode->next = NULL;
    cur->next = nextNode;
}
//弹出栈顶元素
void popTop(struct stack* shead) {
    struct stack* slow = shead;
    struct stack* fast = slow->next;
    if (!fast) return;
    while (fast->next)
    {
        slow = slow->next;
        fast = fast->next;
    }
    free(fast);
    slow->next = NULL;
}
//获取栈顶元素
struct stack* getTop(struct stack* shead) {
    struct stack* cur = shead;
    while (cur->next)
        cur = cur->next;

    return cur;
}


int* preorderTraversal(struct TreeNode* root, int* returnSize) {
    int* ret = (int*)malloc(sizeof(int) * 1000);
    int retIndex = 0;
    struct stack* shead = (struct stack*)malloc(sizeof(struct stack));
    shead->next = NULL;
    struct stack* cur = shead;
    *returnSize = 0;

    if (!root) return ret;

    push_back(root, shead);

    //当栈不为空时
    while (shead->next)
    {
        struct TreeNode* node = getTop(shead)->root;
        ret[retIndex++] = node->val;
        popTop(shead);
        if (node->right) push_back(node->right, shead);
        if (node->left) push_back(node->left, shead);
    }

    *returnSize = retIndex;
    return ret;
}