【练习】输入一颗二元查找树，将该树转换为它的镜像

第15 题：
题目：输入一颗二元查找树，将该树转换为它的镜像，
即在转换后的二元查找树中，左子树的结点都大于右子树的结点。
用递归和循环两种方法完成树的镜像转换。
例如输入：
8
/ \
6 10
/\ /\
5 7 9 11
输出：
8
/ \
10 6
/\ /\
11 9 7 5
定义二元查找树的结点为：
struct BSTreeNode // a node in the binary search tree (BST)
{
	int m_nValue; // value of node
	BSTreeNode *m_pLeft; // left child of node
	BSTreeNode *m_pRight; // right child of node
};
*/

#include <iostream>
using namespace std;

struct BSTreeNode // a node in the binary search tree (BST)
{
	int m_nValue; // value of node
	BSTreeNode *m_pLeft; // left child of node
	BSTreeNode *m_pRight; // right child of node
};

void InsertNode(BSTreeNode* &pRoot,int value)
{
	if (!pRoot)
	{
		pRoot=new BSTreeNode;
		pRoot->m_nValue=value;
		pRoot->m_pLeft=NULL;
		pRoot->m_pRight=NULL;
	} 
	else
	{
		if (pRoot->m_nValue>value)
		{
			InsertNode(pRoot->m_pLeft,value);
		}
		else if (pRoot->m_nValue<value)
		{
			InsertNode(pRoot->m_pRight,value);
		}
		else
		{
			cout<<"repeated insertation"<<endl;
		}
	}

}

void InOrder(BSTreeNode*pRoot)
{
	if (!pRoot)
	{
		return;
	} 
	else
	{
		
		if (pRoot->m_pLeft)
		{
			InOrder(pRoot->m_pLeft);
		}
		cout<<pRoot->m_nValue<<"\t";
		if (pRoot->m_pRight)
		{
			InOrder(pRoot->m_pRight);
		}

	}
}

void Transfer(BSTreeNode* pRoot)
{
	BSTreeNode* pTemp=pRoot->m_pLeft;
	pRoot->m_pLeft=pRoot->m_pRight;
	pRoot->m_pRight=pTemp;

	if (pRoot->m_pLeft)
	{
		Transfer(pRoot->m_pLeft);
	}
	if (pRoot->m_pRight)
	{
		Transfer(pRoot->m_pRight);
	}
	
}


int main()
{
	BSTreeNode *pRoot=NULL;
	InsertNode(pRoot,8);
	InsertNode(pRoot,6);
	InsertNode(pRoot,10);
	InsertNode(pRoot,5);
	InsertNode(pRoot,7);
	InsertNode(pRoot,9);
	InsertNode(pRoot,11);
	cout<<"before transferred"<<endl;
	InOrder(pRoot);
	Transfer(pRoot);
	cout<<"\n after transferred"<<endl;
	InOrder(pRoot);

	return 0;
}