C++在删除BST中只有一个子级的节点时遇到问题

void deleteNode(TreeNode* root, int key) {
    if (root == nullptr) {
        return;
    }
    
    TreeNode* parent = nullptr;
    TreeNode* current = root;
    
    while (current != nullptr && current->val != key) {
        parent = current;
        
        if (key < current->val) {
            current = current->left;
        } else {
            current = current->right;
        }
    }
    
    if (current == nullptr) {
        return;
    }
    
    // Case 1: Node has no child or only one child
    if (current->left == nullptr || current->right == nullptr) {
        TreeNode* child = nullptr;
        
        if (current->left != nullptr) {
            child = current->left;
        } else {
            child = current->right;
        }
        
        if (parent == nullptr) {
            root = child;
        } else if (current == parent->left) {
            parent->left = child;
        } else {
            parent->right = child;
        }
        
        delete current;
    }
    
    // Case 2: Node has two children
    else {
        TreeNode* successor = findSuccessor(current->right);
        current->val = successor->val;
        deleteNode(current->right, successor->val);
    }
}

TreeNode* findSuccessor(TreeNode* node) {
    while (node->left != nullptr) {
        node = node->left;
    }
    
    return node;
}