在一个非空二叉树中统计叶子节点个数。 效果截图： C语言程序代码：

好的，下面是 C 语言代码实现：

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

// 定义二叉树节点结构体
typedef struct TreeNode {
    int val;
    struct TreeNode* left;
    struct TreeNode* right;
} TreeNode;

// 统计叶子节点个数的函数
int count_leaf_node(TreeNode* root) {
    if (!root) {
        return 0;
    }
    if (!root->left && !root->right) {
        return 1;
    }
    return count_leaf_node(root->left) + count_leaf_node(root->right);
}

// 测试
int main() {
    TreeNode* root = (TreeNode*)malloc(sizeof(TreeNode));
    root->val = 1;
    root->left = (TreeNode*)malloc(sizeof(TreeNode));
    root->left->val = 2;
    root->left->left = (TreeNode*)malloc(sizeof(TreeNode));
    root->left->left->val = 4;
    root->left->left->left = NULL;
    root->left->left->right = NULL;
    root->left->right = (TreeNode*)malloc(sizeof(TreeNode));
    root->left->right->val = 5;
    root->left->right->left = NULL;
    root->left->right->right = NULL;
    root->right = (TreeNode*)malloc(sizeof(TreeNode));
    root->right->val = 3;
    root->right->left = (TreeNode*)malloc(sizeof(TreeNode));
    root->right->left->val = 6;
    root->right->left->left = NULL;
    root->right->left->right = NULL;
    root->right->right = NULL;
    printf("%d\n", count_leaf_node(root)); // 输出 3
    return 0;
}

希望能够解决你的问题！