今天去健身房锻炼了身体

这是关于二叉树如何求度为一的节点的个数，同理还能求度为零和二的，不难。
还又复习了一遍前序中序后续的遍历方法，已经可以由任意两种推出二叉树结构了，不过二叉树的样子和模式我还是有点不太能和代码结合去理解，还需要多加练习

include <stdio.h>

include <stdlib.h>

typedef struct TreeNode {
int value;
struct TreeNode *left;
struct TreeNode *right;
} TreeNode;

TreeNode *createTreeNode(int value) {
TreeNode *node = (TreeNode *)malloc(sizeof(TreeNode));
node->value = value;
node->left = NULL;
node->right = NULL;
return node;
}

int countDegreeOneNodes(TreeNode *root) {
if (root == NULL) {
return 0;
}
int count = 0;
if ((root->left!= NULL && root->right == NULL) || (root->left == NULL && root->right!= NULL)) {
count++;
}
return count + countDegreeOneNodes(root->left) + countDegreeOneNodes(root->right);
}

int main() {
TreeNode *root = createTreeNode(1);
root->left = createTreeNode(2);
root->right = createTreeNode(3);
root->left->right = createTreeNode(4);

int result = countDegreeOneNodes(root);
printf("度为 1 的节点个数：%d\n", result);

return 0;

}