c语言编程。用“孩子链表表示法”将每个结点的孩子结点链接构成一个单链表，称为孩子链表。将孩子链表的头指针又组成一个顺序表。设计函数将二叉树以“孩子链表表示法”输入到计算机中。设计函数计算“孩子链表表示法”下二叉树的叶子结点个数。

好的，我会尽力回答。下面是 C 语言的代码实现。 首先，我们需要定义一个二叉树结构体，其中包含左子树指针、右子树指针和数据域。 ```c typedef struct node { int data; struct node *left; struct node *right; } Node; ``` 接下来，我们通过孩子链表表示法将二叉树输入到计算机中。我们需要定义一个结构体来表示孩子链表节点，其中包含孩子节点指针和数据域。 ```c typedef struct child_node { int data; struct child_node *next; } ChildNode; typedef struct child_list { ChildNode *head; } ChildList; ``` 在输入二叉树时，我们需要使用递归函数，每次将左子树和右子树分别插入到孩子链表中，最后将孩子链表头指针插入到顺序表中。 ```c void input_tree(Node *tree, ChildList *clist) { if (tree == NULL) { return; } ChildNode *new_child_node = (ChildNode *)malloc(sizeof(ChildNode)); new_child_node->data = tree->data; new_child_node->next = NULL; if (clist->head == NULL) { clist->head = new_child_node; } else { ChildNode *cur = clist->head; while (cur->next != NULL) { cur = cur->next; } cur->next = new_child_node; } input_tree(tree->left, clist); input_tree(tree->right, clist); } ``` 最后，我们需要计算孩子链表表示法下二叉树的叶子结点个数。我们可以使用递归函数遍历孩子链表，如果当前节点没有孩子节点，说明它是一个叶子节点。 ```c int count_leaves(ChildNode *head) { if (head == NULL) { return 0; } if (head->next == NULL) { return 1; } int count = 0; ChildNode *cur = head; while (cur != NULL) { count += count_leaves(cur->next); cur = cur->next; } return count; } ``` 完整代码如下： ```c #include <stdio.h> #include <stdlib.h> typedef struct node { int data; struct node *left; struct node *right; } Node; typedef struct child_node { int data; struct child_node *next; } ChildNode; typedef struct child_list { ChildNode *head; } ChildList; void input_tree(Node *tree, ChildList *clist); int count_leaves(ChildNode *head); int main() { // 构造一棵二叉树 Node *tree = (Node *)malloc(sizeof(Node)); tree->data = 1; tree->left = (Node *)malloc(sizeof(Node)); tree->left->data = 2; tree->left->left = NULL; tree->left->right = NULL; tree->right = (Node *)malloc(sizeof(Node)); tree->right->data = 3; tree->right->left = (Node *)malloc(sizeof(Node)); tree->right->left->data = 4; tree->right->left->left = NULL; tree->right->left->right = NULL; tree->right->right = (Node *)malloc(sizeof(Node)); tree->right->right->data = 5; tree->right->right->left = NULL; tree->right->right->right = NULL; // 将二叉树以孩子链表表示法输入到计算机中 ChildList *clist = (ChildList *)malloc(sizeof(ChildList)); clist->head = NULL; input_tree(tree, clist); // 计算叶子结点个数 int count = count_leaves(clist->head); printf("叶子结点个数为：%d\n", count); return 0; } void input_tree(Node *tree, ChildList *clist) { if (tree == NULL) { return; } ChildNode *new_child_node = (ChildNode *)malloc(sizeof(ChildNode)); new_child_node->data = tree->data; new_child_node->next = NULL; if (clist->head == NULL) { clist->head = new_child_node; } else { ChildNode *cur = clist->head; while (cur->next != NULL) { cur = cur->next; } cur->next = new_child_node; } input_tree(tree->left, clist); input_tree(tree->right, clist); } int count_leaves(ChildNode *head) { if (head == NULL) { return 0; } if (head->next == NULL) { return 1; } int count = 0; ChildNode *cur = head; while (cur != NULL) { count += count_leaves(cur->next); cur = cur->next; } return count; } ```