1.用非递归遍历求二叉树结点个数【计学2020】

算法思想:用先序非递归遍历

当前指针不为空或栈不为空进行循环；

当前指针不为空访问当前结点，当前节点入栈，进入左子树

当前指针为空，栈顶元素出栈（回溯），进入右子树

#include<malloc.h>
#define MaxSize 100
typedef struct node
{
	int data;
	struct node* left;
	struct node* right;
}BTNode;

BTNode* BuyNode(int a)//a是要创建的值
{
	BTNode* newnode = (BTNode*)malloc(sizeof(BTNode));//定义临时变量扩容
	if (newnode == NULL)
	{
		printf("malloc fail");//扩容失败
		return NULL;
	}
	newnode->data = a;
	newnode->left = NULL;
	newnode->right = NULL;
	return newnode;
}


BTNode* BinaryTreeCreate()//二叉树一般需要手动扩建
{
	BTNode* n1 = BuyNode(5);
	BTNode* n2 = BuyNode(4);
	BTNode* n3 = BuyNode(6);
	BTNode* n4 = BuyNode(2);
	BTNode* n5 = BuyNode(1);

	n1->left = n2;
	n1->right = n3;
	n2->left = n4;
	n2->right = n5;

	return n1;
}

int preOrder(BTNode* T)
{
	BTNode* stack[MaxSize];
	int count = 0;//节点个数
	int top = -1;//栈顶指针
	BTNode* cur = T;
	while(cur != NULL || top != -1)
	{
		if(cur != NULL)
		{
			count++;//相当于根 ,计数器 
			stack[++top] = cur;//入栈 
			cur = cur->left;//遍历左子树 
		}
		else
		{
			cur = stack[top];
			top--;//栈顶元素出栈 
			cur = cur->right;//遍历右子树 
		}
	}
	return count;
}
// 二叉树前序遍历 
void BinaryTreePrevOrder(BTNode* root) 
{
	if (root == NULL)
	{

		return;
	}
	printf("%d ", root->data);//打印节点
	BinaryTreePrevOrder(root->left);//访问左节点
	BinaryTreePrevOrder(root->right);//访问右节点

}
int main()
{

	BTNode* T = BinaryTreeCreate();//创建二叉树

	printf("%d",preOrder(T));
	BinaryTreePrevOrder(T);
	
	return 0;
}

2.判断二叉树中是否有平衡结点

算法思想：

 递归三部曲

  1、找递归终止条件
  2、找返回值：应该给上一级返回什么信息？
  3、本级递归应该做什么：在这一级递归中，应该完成什任务？

        本级递归需要算出左子树之和与右子树之和并比较

int postOrder(BTNode* root,bool* have)
{
	if(root == NULL)//
		return 0;
	if(root->left == NULL && root->right == NULL)//递归终止条件 
	{
		return root->data;
	}
	int leftSum =  postOrder(root->left, have);
	int rightSum =  postOrder(root->right, have);
	if(leftSum == rightSum)
	{
		*have = true;
	}
	return leftSum + rightSum + root->data;//返回给上一层的信息：算出左右子树以及当前节点之和
}
int isBalance(BTNode* root)
{
	bool have = false;
	postOrder(root,&have);
	return have;
}