以二叉链表作为二叉树的存储结构，编写用层次顺序遍历二叉树的方法，统计树中度为1 的结点个数。

以下是用C++实现的代码：

#include <iostream>
#include <queue>
using namespace std;

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

int countDegreeOne(TreeNode* root) {
    if (!root) return 0;
    queue<TreeNode*> q;
    q.push(root);
    int count = 0;
    while (!q.empty()) {
        TreeNode* curr = q.front();
        q.pop();
        if (curr->left && !curr->right || !curr->left && curr->right) {
            count++;
        }
        if (curr->left) {
            q.push(curr->left);
        }
        if (curr->right) {
            q.push(curr->right);
        }
    }
    return count;
}

int main() {
    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);

    int count = countDegreeOne(root);
    cout << "The number of degree one nodes in the tree is: " << count << endl;

    return 0;
}

该算法的时间复杂度为O(n)，其中n是二叉树中的结点数。