用c语言写份求二叉树中的（叶子/单分支/双分支）结点个数的代码

以下是用C语言实现求二叉树中叶子、单分支、双分支结点个数的代码：

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

// 结点结构体
typedef struct TreeNode {
    int val;
    struct TreeNode* left;
    struct TreeNode* right;
} TreeNode;

// 构造二叉树
TreeNode* createTree() {
    int val;
    scanf("%d", &val);
    if (val == -1) {
        return NULL;
    }
    TreeNode* root = (TreeNode*)malloc(sizeof(TreeNode));
    root->val = val;
    root->left = createTree();
    root->right = createTree();
    return root;
}

// 计算叶子结点个数
int countLeafNodes(TreeNode* root) {
    if (root == NULL) {
        return 0;
    }
    if (root->left == NULL && root->right == NULL) {
        return 1;
    }
    return countLeafNodes(root->left) + countLeafNodes(root->right);
}

// 计算单分支结点个数
int countSingleBranchNodes(TreeNode* root) {
    if (root == NULL) {
        return 0;
    }
    if (root->left == NULL && root->right == NULL) {
        return 0;
    }
    if (root->left == NULL && root->right != NULL) {
        return 1 + countSingleBranchNodes(root->right);
    }
    if (root->left != NULL && root->right == NULL) {
        return 1 + countSingleBranchNodes(root->left);
    }
    return countSingleBranchNodes(root->left) + countSingleBranchNodes(root->right);
}

// 计算双分支结点个数
int countDoubleBranchNodes(TreeNode* root) {
    if (root == NULL) {
        return 0;
    }
    if (root->left == NULL && root->right == NULL) {
        return 0;
    }
    if (root->left != NULL && root->right != NULL) {
        return 1 + countDoubleBranchNodes(root->left) + countDoubleBranchNodes(root->right);
    }
    return countDoubleBranchNodes(root->left) + countDoubleBranchNodes(root->right);
}

int main() {
    TreeNode* root = createTree();
    printf("叶子结点个数：%d\n", countLeafNodes(root));
    printf("单分支结点个数：%d\n", countSingleBranchNodes(root));
    printf("双分支结点个数：%d\n", countDoubleBranchNodes(root));
    return 0;
}

这里我们通过递归的方式来实现计算各种类型结点的个数。其中，countLeafNodes函数用来计算叶子结点个数，countSingleBranchNodes函数用来计算单分支结点个数，countDoubleBranchNodes函数用来计算双分支结点个数。在计算单分支结点和双分支结点的时候，我们通过判断结点的左右子树是否为空来进行区分。