用c++编写出以先序遍历的顺序建立二叉树，并输出度为2 的结点数。

好的，根据您的要求，我来为您解答。

首先，先给出二叉树的结构体定义：

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

然后，给出以先序遍历的顺序建立二叉树的函数：

TreeNode* buildTree(vector<int>& preorder) {
    if (preorder.empty()) {
        return nullptr;
    }
    int val = preorder[0];
    TreeNode* root = new TreeNode(val);
    vector<int> left_preorder, right_preorder;
    for (int i = 1; i < preorder.size(); i++) {
        if (preorder[i] < val) {
            left_preorder.push_back(preorder[i]);
        } else {
            right_preorder.push_back(preorder[i]);
        }
    }
    root->left = buildTree(left_preorder);
    root->right = buildTree(right_preorder);
    return root;
}

最后，给出统计度为2的节点数的函数：

int countDegree2(TreeNode* root) {
    if (root == nullptr) {
        return 0;
    }
    int count = 0;
    if (root->left != nullptr && root->right != nullptr) {
        count++;
    }
    count += countDegree2(root->left);
    count += countDegree2(root->right);
    return count;
}

完整代码如下：

#include <iostream>
#include <vector>

using namespace std;

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

TreeNode* buildTree(vector<int>& preorder) {
    if (preorder.empty()) {
        return nullptr;
    }
    int val = preorder[0];
    TreeNode* root = new TreeNode(val);
    vector<int> left_preorder, right_preorder;
    for (int i = 1; i < preorder.size(); i++) {
        if (preorder[i] < val) {
            left_preorder.push_back(preorder[i]);
        } else {
            right_preorder.push_back(preorder[i]);
        }
    }
    root->left = buildTree(left_preorder);
    root->right = buildTree(right_preorder);
    return root;
}

int countDegree2(TreeNode* root) {
    if (root == nullptr) {
        return 0;
    }
    int count = 0;
    if (root->left != nullptr && root->right != nullptr) {
        count++;
    }
    count += countDegree2(root->left);
    count += countDegree2(root->right);
    return count;
}

int main() {
    vector<int> preorder = {5, 3, 2, 4, 7, 6, 8};
    TreeNode* root = buildTree(preorder);
    int count = countDegree2(root);
    cout << "度为2的结点数为：" << count << endl;
    return 0;
}

希望能够帮到您！