🖊作者 : D. Star.
📘专栏 : 数据结构
😆今日分享 : 为什么阴干的衣服会有味道?
衣服上超过三分之二的污垢来自我们身体外部的分泌物，即汗水和皮脂。如果在洗衣服的过程中皮脂没有洗干净的话，它就可以成为衣服（没有正确晒干）臭味来源的罪魁祸首。因为这些天然的油没有洗掉就会沉到布料的纤维中，并且它们非常粘稠，容易吸引污垢。这就为细菌真菌的繁殖提供了一个绝佳的“培养皿”。如果天气阴冷潮湿，无法让衣服晒干，细菌会跟衣服上残留的汗液、皮肤细胞和其他各种腺体分泌物发生反应，产生挥发性有机化合物，就会产生臭味。

🌻头文件

#define _CRT_SECURE_NO_WARNINGS
#pragma once
#include<stdio.h>
#include<stdlib.h>


#define DataType int

typedef struct tree
{
	DataType a;
	struct tree* left;
	struct tree* right;
}tree;


//申请一个节点
tree* BuyNode(DataType x);

//二叉树:
//前序遍历
void PreOrder(tree* t);
//中序遍历
void InOrder(tree* t);
//后序遍历
void PostOrder(tree* t);


//求二叉树的节点数1
int TreeSize1(tree* t);
//求二叉树的节点数2
int TreeSize2(tree* t);

//叶子节点数
int TreeLeafSize(tree* t);

//求二叉树的高度
int TreeHeight(tree* t);

//求k层上的节点个数  k >=1	
int TreeLevelSize(tree* t,int k);

🌻功能实现(详细版)

🔎申请结点

//申请一个节点
tree* BuyNode(DataType x)
{
	tree* d = (tree*)malloc(sizeof(tree));
	if (d == NULL)
	{
		perror("malloc fail");
		exit(-1);
	}
	d->a = x;
	d->left = d->right = NULL;
	return d;
}

🔎递归算法:

✔前序遍历

//二叉树的前序遍历
void PreOrder(tree* t)
{
	if (t == NULL)
	{
		printf("# ");
		return;
	}
	else 
	{
		printf("%d ",t->a);
		PreOrder(t->left);
		PreOrder(t->right);
	}
}

✔中序遍历

//中序遍历
void InOrder(tree* t)
{
	if (t == NULL)
	{
		printf("\n");
		return;
	}
	else
	{
		PreOrder(t->left);
		printf("%d ", t->a);
		PreOrder(t->right);
	}
}

✔后序遍历

//后序遍历
void PostOrder(tree* t)
{
	if (t == NULL)
	{
		printf("\n");
		return;
	}
	else
	{
		PreOrder(t->left);
		PreOrder(t->right);
		printf("%d ", t->a);
	}
}

🔎求二叉树的总结点数(法一/法二)

int size = 0;
//求二叉树的节点数1
int TreeSize1(tree* t)
{
	if (t == NULL)
		return;

	size++;
	TreeSize1(t->left);
	TreeSize1(t->right);

	return size;
}

//求二叉树的节点数2
int TreeSize2(tree* t)
{
	return t == NULL ? 0 :
		TreeSize2(t->left) + TreeSize2(t->right) + 1;
}

🔎二叉树的叶子节点数

//叶子节点数
int TreeLeafSize(tree* t)
{
	if (t == NULL)
	{
		return 0;
	}
	if (t->left == NULL && t->right == NULL)
	{
		return 1;
	}
	return TreeLeafSize(t->left) + TreeLeafSize(t->right);
}

🔎求二叉树的高度

//求二叉树的高度
int TreeHeight(tree* t)
{
	if (t == NULL)
		return 0;
	int leftT = TreeHeight(t->left);
	int rightT = TreeHeight(t->right);
	return (leftT > rightT ? leftT : rightT) + 1;
}

🔎求k层上的结点个数(k>=1)

//求k层上的节点个数  k >=1	
int TreeLevelSize(tree* t, int k)
{
	if (t == NULL)
		return 0;
	if (k == 1)
	{
		return 1;
	}
	return TreeLevelSize(t->left, k - 1) + TreeLevelSize(t->right, k - 1);
}

🌻功能实现(全部代码)

#define _CRT_SECURE_NO_WARNINGS
#include"tree.h"

//申请一个节点
tree* BuyNode(DataType x)
{
	tree* d = (tree*)malloc(sizeof(tree));
	if (d == NULL)
	{
		perror("malloc fail");
		exit(-1);
	}
	d->a = x;
	d->left = d->right = NULL;
	return d;
}
//****************递归算法*******************************

//二叉树的前序遍历
void PreOrder(tree* t)
{
	if (t == NULL)
	{
		printf("# ");
		return;
	}
	else 
	{
		printf("%d ",t->a);
		PreOrder(t->left);
		PreOrder(t->right);
	}
}

//中序遍历
void InOrder(tree* t)
{
	if (t == NULL)
	{
		printf("\n");
		return;
	}
	else
	{
		PreOrder(t->left);
		printf("%d ", t->a);
		PreOrder(t->right);
	}
}

//后序遍历
void PostOrder(tree* t)
{
	if (t == NULL)
	{
		printf("\n");
		return;
	}
	else
	{
		PreOrder(t->left);
		PreOrder(t->right);
		printf("%d ", t->a);
	}
}

int size = 0;
//求二叉树的节点数
int TreeSize1(tree* t)
{
	if (t == NULL)
		return;

	size++;
	TreeSize1(t->left);
	TreeSize1(t->right);

	return size;
}

//求二叉树的节点数
int TreeSize2(tree* t)
{
	return t == NULL ? 0 :
		TreeSize2(t->left) + TreeSize2(t->right) + 1;
}

//叶子节点数
int TreeLeafSize(tree* t)
{
	if (t == NULL)
	{
		return 0;
	}
	if (t->left == NULL && t->right == NULL)
	{
		return 1;
	}
	return TreeLeafSize(t->left) + TreeLeafSize(t->right);
}

//求二叉树的高度
int TreeHeight(tree* t)
{
	if (t == NULL)
		return 0;
	int leftT = TreeHeight(t->left);
	int rightT = TreeHeight(t->right);
	return (leftT > rightT ? leftT : rightT) + 1;
}

//求k层上的节点个数  k >=1	
int TreeLevelSize(tree* t, int k)
{
	if (t == NULL)
		return 0;
	if (k == 1)
	{
		return 1;
	}
	return TreeLevelSize(t->left, k - 1) + TreeLevelSize(t->right, k - 1);
}

🌻测试代码

#define _CRT_SECURE_NO_WARNINGS
#include"tree.h"


test01()
{
	tree* n1 = BuyNode(1);
	tree* n2 = BuyNode(2);
	tree* n3 = BuyNode(3);
	tree* n4 = BuyNode(4);
	tree* n5 = BuyNode(5);
	tree* n6 = BuyNode(6);
	n1->left = n2;
	n2->left = n4;
	n2->right = n5;
	n1->right = n3;
	n3->left = n6;
	printf("二叉树的前序遍历结果为：");
	PreOrder(n1);
	printf("\n");

	
	printf("结点个数为：%d\n", TreeSize1(n1));
	printf("结点个数为：%d\n",TreeSize2(n1));
	printf("叶子节点个数为：%d\n",TreeLeafSize(n1));
	printf("二叉树的高度为：%d\n",TreeHeight(n1));
	printf("第k层上的节点数为：%d\n", TreeLevelSize(n1, 2));
}

int main()
{
	test01();
	system("pause");
	return 0;
}

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