C++：如何计算二叉树中其值模块的高度小于2的节点数？

struct TreeNode {
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};

int countNodes(TreeNode* root) {
    if (root == nullptr) {
        return 0;
    }
    
    int count = 0;
    if (root->val % 2 < 2) {
        count++;
    }
    
    count += countNodes(root->left);
    count += countNodes(root->right);
    
    return count;
}

int main() {
    // 创建二叉树
    TreeNode* root = new TreeNode(1);
    root->left = new TreeNode(2);
    root->right = new TreeNode(3);
    root->left->left = new TreeNode(4);
    root->left->right = new TreeNode(5);
    root->right->left = new TreeNode(6);
    root->right->right = new TreeNode(7);
    
    // 计算满足条件的节点数
    int result = countNodes(root);
    cout << "满足条件的节点数：" << result << endl;
    
    // 释放内存
    delete root->left->left;
    delete root->left->right;
    delete root->right->left;
    delete root->right->right;
    delete root->left;
    delete root->right;
    delete root;
    
    return 0;
}