求二元查找树的镜像

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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 nodeBSTreeNode  *m_pLeft;  // left child of nodeBSTreeNode  *m_pRight; // right child of node};void create(BSTreeNode *&p,int *a,int i){if(i>7)return;p=new BSTreeNode;p->m_nValue=a[i];p->m_pLeft=p->m_pRight=NULL;create(p->m_pLeft,a,i*2);create(p->m_pRight,a,i*2+1);}void Turn(BSTreeNode *&p){if(p==NULL)return;BSTreeNode *temp;temp=p->m_pLeft;p->m_pLeft=p->m_pRight;p->m_pRight=temp;Turn(p->m_pLeft);Turn(p->m_pRight);}int main(){BSTreeNode *p;int a[]={0,8,6,10,5,7,9,11};create(p,a,1);Turn(p);return 0;}

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