4.29 数据结构 二叉树

手动输入节点值，二叉树的前序、中序、后序遍历及销毁 并计算二叉树的节点数

运行结果截图:

代码：

/*许沐 20760204
4.28 二叉树 */

#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<string.h>   
#include<iostream>
#include<stdlib.h>
using namespace std;
#define MAXSIZE 100


typedef int SqBiTree[MAXSIZE];
SqBiTree bt;

typedef struct BTNode
{
	int data;
	struct BTNode* lChild;
	struct BTNode* rChild;
}BiTNode;

void CreateBiTree(BiTNode** T)
{
	int ch;
	scanf_s("%d", &ch);
	if (ch < 0)
	{
		*T = NULL;
		return;
	}
	else
	{
		*T = (BiTNode*)malloc(sizeof(BiTNode));
		(*T)->data = ch;
		printf("请输入%d的左子节点：", ch);
		CreateBiTree(&((*T)->lChild));
		printf("请输入%d的右子节点：", ch);
		CreateBiTree(&((*T)->rChild));
	}
	return;
}

void PreOrderBiTree(BiTNode* T) //先序遍历二叉树
{
	if (T == NULL)
		return;
	else
	{
		printf("%2d", T->data);
		PreOrderBiTree(T->lChild);
		PreOrderBiTree(T->lChild);
	}
}

void MiddleOrderBiTree(BiTNode* T)  //中序遍历二叉树
{
	if (T == NULL)
		return;
	else
	{
		MiddleOrderBiTree(T->lChild);
		printf("%2d", T->data);
		MiddleOrderBiTree(T->rChild);
	}
}

void PostOrderBiTree(BiTNode* T)  //后序遍历二叉树
{
	if (T == NULL)
		return;
	else
	{
		PostOrderBiTree(T->lChild);		
		PostOrderBiTree(T->rChild);
		printf("%2d", T->data);
	}
}

int LeafCount(BiTNode* T)  //计算叶子节点个数
{
	static int l = 0;
	if (T != NULL)
	{
		if (T->lChild == NULL && T->rChild == NULL)
			l++;
		LeafCount(T->lChild);
		LeafCount(T->rChild);
	}
	return l;
}

void FreeTree(BiTNode* T) //二叉树的销毁
{
	if (T!=NULL)
	{
		if (T->lChild)
		{
			FreeTree(T->lChild);
			T->lChild = NULL;
		}
		if (T->rChild)
		{
			FreeTree(T->rChild);
			T->rChild = NULL;
		}
		if (T != NULL)
		{
			free(T);
			T = NULL;
		}
	}
}


int main()
{
	BiTNode* T;
	T = NULL;
	int nLeafCount = 0;
	printf("请输入第一个节点的值，（-1表示没有叶节点）:\n");
	CreateBiTree(&T);

	printf("先序遍历二叉树：");
	PreOrderBiTree(T);
	printf("\n");

	printf("中序遍历二叉树：");
	MiddleOrderBiTree(T);
	printf("\n");

	printf("后序遍历二叉树：");
	PostOrderBiTree(T);
	printf("\n");

	nLeafCount = LeafCount(T);;
	printf("叶子节点的个数为%d", nLeafCount);

	FreeTree(T);

	return 0;
}