6-1 求二叉树高度(20 分)

6-1 求二叉树高度(20 分)

本题要求给定二叉树的高度。

函数接口定义:

int GetHeight( BinTree BT );

其中BinTree结构定义如下:

typedef struct TNode *Position;
typedef Position BinTree;
struct TNode{
    ElementType Data;
    BinTree Left;
    BinTree Right;
};

要求函数返回给定二叉树BT的高度值。

裁判测试程序样例:

#include <stdio.h>
#include <stdlib.h>

typedef char ElementType;
typedef struct TNode *Position;
typedef Position BinTree;
struct TNode{
    ElementType Data;
    BinTree Left;
    BinTree Right;
};

BinTree CreatBinTree(); /* 实现细节忽略 */
int GetHeight( BinTree BT );

int main()
{
    BinTree BT = CreatBinTree();
    printf("%d\n", GetHeight(BT));
    return 0;
}
/* 你的代码将被嵌在这里 */

输出样例(对于图中给出的树):

4
#include <stdio.h>
#include <stdlib.h>

typedef char ElementType;
typedef struct TNode *Position;
typedef Position BinTree;
struct TNode {//结点结构
	ElementType Data;
	BinTree Left;
	BinTree Right;
};

BinTree CreatBinTree(); //创建树,题目不要求实现,为了理解,我还是实现了他
int GetHeight(BinTree BT);//递归输出

int main(void)
{
	BinTree BT = CreatBinTree();
	printf("%d\n", GetHeight(BT));

	return 0;
}
BinTree CreatBinTree()
{//层序生成二叉树
	//特殊限制,输入为0的时候为叶节点
	BinTree BT;
	ElementType T;
	int front = 0, tail = 0;
	BinTree Queue[1001] = { '\0' };//父节点数列
	BinTree Date;
	scanf("%c", &T);
	if (T == '0')//空树
		return NULL;
	else
	{
		BT = (BinTree)malloc(sizeof(struct TNode));
		if (BT == NULL)
			return NULL;
		BT->Data = T;//根节点赋值
		BT->Left = BT->Right = NULL;//初始化左右子树
		Queue[tail++] = BT;
	}
	while (Queue[front]!=NULL) {//根节点入队列
		Date = Queue[front++];
		scanf("%c", &T);
		if (T == '0')
			Date->Left = NULL;
		else
		{
			Date->Left = (BinTree)malloc(sizeof(struct TNode));
			if (Date->Left == NULL)//先完成左树的操作,然后完成右树的操作
				return NULL;
			Date->Left->Data = T;
			Date->Left->Left = Date->Left->Right = NULL;
			Queue[tail++] = Date->Left;
		}
		scanf("%c", &T);
		if (T == '0')
			Date->Right = NULL;
		else
		{
			Date->Right = (BinTree)malloc(sizeof(struct TNode));
			if (Date->Right == NULL)
				return NULL;
			Date->Right->Data = T;
			Date->Right->Left = Date->Right->Right = NULL;
			Queue[tail++] = Date->Right;
		}
	};

	return BT;
}
int GetHeight(BinTree BT)//递归方法
{
	int LH, RH;//对左右子树的高度进行记录
	if (!BT)//末位,递归停止
		return 0;
	else
	{
		LH = GetHeight(BT->Left);
		RH = GetHeight(BT->Right);
		return LH > RH ? ++LH : ++RH;//返回左右子树中值最大的,再加上当前父节点,因为不会统计父节点
	}
}

猜你喜欢

转载自blog.csdn.net/mathew_leung/article/details/79798593
今日推荐