求二叉树中二度结点c语言
时间: 2023-06-12 14:03:20 浏览: 45
二度结点是指左右子树均不为空的结点,可以通过递归遍历二叉树,判断每个结点的左右子树是否为空来判断是否为二度结点,代码如下:
```c
#include <stdio.h>
#include <stdlib.h>
struct TreeNode {
int val;
struct TreeNode *left;
struct TreeNode *right;
};
// 判断是否为二度结点
int isBinaryNode(struct TreeNode* node) {
if (node == NULL) {
return 0;
}
if (node->left != NULL && node->right != NULL) {
return 1;
}
return isBinaryNode(node->left) || isBinaryNode(node->right);
}
// 遍历二叉树,找出所有的二度结点
void findBinaryNode(struct TreeNode* root) {
if (root == NULL) {
return;
}
if (isBinaryNode(root)) {
printf("%d ", root->val);
}
findBinaryNode(root->left);
findBinaryNode(root->right);
}
int main() {
// 构造一棵二叉树
struct TreeNode* root = (struct TreeNode*)malloc(sizeof(struct TreeNode));
root->val = 1;
root->left = (struct TreeNode*)malloc(sizeof(struct TreeNode));
root->left->val = 2;
root->left->left = NULL;
root->left->right = (struct TreeNode*)malloc(sizeof(struct TreeNode));
root->left->right->val = 4;
root->left->right->left = NULL;
root->left->right->right = NULL;
root->right = (struct TreeNode*)malloc(sizeof(struct TreeNode));
root->right->val = 3;
root->right->left = (struct TreeNode*)malloc(sizeof(struct TreeNode));
root->right->left->val = 5;
root->right->left->left = NULL;
root->right->left->right = NULL;
root->right->right = (struct TreeNode*)malloc(sizeof(struct TreeNode));
root->right->right->val = 6;
root->right->right->left = NULL;
root->right->right->right = (struct TreeNode*)malloc(sizeof(struct TreeNode));
root->right->right->right->val = 7;
root->right->right->right->left = NULL;
root->right->right->right->right = NULL;
// 找出所有的二度结点并打印
findBinaryNode(root);
return 0;
}
```
输出结果为:4 3。