C++ 合并2个二叉树时出现运行时错误

我正试图从leetcode解决这个问题。这是我的C++代码< /p>

    class Solution {
public:
    TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
        TreeNode *t;
        if(t1==nullptr){return t2;}
        if(t2==nullptr){return t1;}
        t->val=t1->val+t2->val;
        t->left=mergeTrees(t1->left,t2->left);
        t->right=mergeTrees(t1->right,t2->right);
        return t;
    }
};

此代码产生以下错误

Line 18: Char 12: runtime error: member access within misaligned address 0x000000000001 for type 'TreeNode', which requires 8 byte alignment (solution.cpp)
0x000000000001: note: pointer points here
<memory cannot be printed>
SUMMARY: UndefinedBehaviorSanitizer: undefined-behavior prog_joined.cpp:27:12

有人能解释一下为什么第一个代码会导致错误吗？

TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
    TreeNode *t;
    if(t1==nullptr){return t2;}
    if(t2==nullptr){return t1;}
    t->val=t1->val+t2->val;

t是未初始化的指针，因此t->val是一个错误

你大概是指这样的事吧

TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
    if(t1==nullptr){return t2;}
    if(t2==nullptr){return t1;}
    TreeNode *t = new TreeNode;
    t->val=t1->val+t2->val;

它确实解决了这个问题。 然而，我们仍然可以在这里简化我们的陈述：

---

您将“t”初始化为指向树节点的指针，然后应该添加t=new树节点；在您编写TreeNode*t；之后；。希望能奏效：

看起来我们需要TreeNode类或结构的声明和定义。好吧，我们需要一个更精确的定义。@ThomasMatthews该链接在右侧显示了TreeNode的定义side@MonkeyDLuffy：抱歉，防火墙阻止我访问链接。

TreeNode* mergeTrees(TreeNode* t1, TreeNode* t2) {
    if(t1==nullptr){return t2;}
    if(t2==nullptr){return t1;}
    TreeNode *t = new TreeNode;
    t->val=t1->val+t2->val;

// The following block might slightly improve the execution time;
// Can be removed;
static const auto __optimize__ = []() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    std::cout.tie(nullptr);
    return 0;
}();

static const struct Solution {
    TreeNode* mergeTrees(
        TreeNode* t1,
        TreeNode* t2
    ) {
        if (t1 && t2) {
            TreeNode* root = new TreeNode(t1->val + t2->val);
            root->left = mergeTrees(t1->left, t2->left);
            root->right = mergeTrees(t1->right, t2->right);
            return root;
        }
        
        return t1 ? t1 : t2;
    }
};