假设二叉树中每个结点值为单个字符，采用二叉链存储结构存储设计，一个算法计算一棵树给定二叉树BT中的所有单分支结点个数.用c++

好的，下面是该算法的C++代码实现：

#include <iostream>
using namespace std;

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

int countSingleBranchNodes(TreeNode* root) {
    if (!root) {
        return 0;
    }
    
    int count = 0;
    
    if (root->left && !root->right) {
        count += 1;
    } else if (root->right && !root->left) {
        count += 1;
    }
    
    count += countSingleBranchNodes(root->left) + countSingleBranchNodes(root->right);
    
    return count;
}

int main() {
    // 构造一棵二叉树
    TreeNode* root = new TreeNode('A');
    root->left = new TreeNode('B');
    root->right = new TreeNode('C');
    root->left->left = new TreeNode('D');
    root->left->right = new TreeNode('E');
    root->right->left = new TreeNode('F');
    root->right->right = new TreeNode('G');
    root->left->left->left = new TreeNode('H');
    root->left->left->right = new TreeNode('I');
    root->left->right->left = new TreeNode('J');
    root->right->right->left = new TreeNode('K');
    root->right->right->right = new TreeNode('L');
    
    int count = countSingleBranchNodes(root);
    
    cout << "单分支结点的个数为：" << count << endl;
    
    return 0;
}

这个算法的时间复杂度是O(n)，其中n是二叉树中节点的数量。