以二叉链表作为二叉树的存储结构， 编写以下算法： (l) 统计二叉树的叶结点个数。用c语言写出完整的代码与注释

以下是统计二叉树叶节点个数的C语言代码及注释：

#include <stdio.h>
#include <stdlib.h>

// 定义二叉树节点结构体
typedef struct TreeNode {
    int val; // 节点值
    struct TreeNode *left; // 左子节点指针
    struct TreeNode *right; // 右子节点指针
} TreeNode;

// 统计二叉树叶节点个数的函数
int countLeaves(TreeNode* root) {
    if (!root) { // 如果当前节点为空，返回0
        return 0;
    } else if (!root->left && !root->right) { // 如果当前节点为叶节点，返回1
        return 1;
    } else { // 否则递归遍历左右子树，并将结果相加
        return countLeaves(root->left) + countLeaves(root->right);
    }
}

int main() {
    // 构建以下二叉树
    //     1
    //    / \
    //   2   3
    //  / \
    // 4   5
    TreeNode* root = (TreeNode*)malloc(sizeof(TreeNode));
    root->val = 1;
    root->left = (TreeNode*)malloc(sizeof(TreeNode));
    root->right = (TreeNode*)malloc(sizeof(TreeNode));
    root->left->val = 2;
    root->right->val = 3;
    root->left->left = (TreeNode*)malloc(sizeof(TreeNode));
    root->left->right = (TreeNode*)malloc(sizeof(TreeNode));
    root->left->left->val = 4;
    root->left->right->val = 5;
    root->left->left->left = NULL;
    root->left->left->right = NULL;
    root->left->right->left = NULL;
    root->left->right->right = NULL;
    root->right->left = NULL;
    root->right->right = NULL;

    // 统计叶节点个数并输出
    int count = countLeaves(root);
    printf("The number of leaves in the binary tree is %d.\n", count);

    return 0;
}

运行结果：

The number of leaves in the binary tree is 3.