把二元查找树转变成排序的双向链表

题目：
输入一棵二元查找树，将该二元查找树转换成一个排序的双向链表。
要求不能创建任何新的结点，只调整指针的指向。
10
/ \
6 14
/ \ / \
4 8 12 16
转换成双向链表
4=6=8=10=12=14=16。
首先我们定义的二元查找树节点的数据结构如下：
struct BSTreeNode
{
int m_nValue; // value of node
BSTreeNode *m_pLeft; // left child of node
BSTreeNode *m_pRight; // right child of node

};

#include <iostream>
#include <vector>

using namespace std;

struct BSTreeNode
{
 int m_nValue;
 BSTreeNode *m_pLeft;
 BSTreeNode *m_pRight;
};


BSTreeNode* createNode(int num)
{
 BSTreeNode* node = (BSTreeNode*)malloc(sizeof(BSTreeNode));

 if(!node)
  return NULL;
 node->m_nValue = num;
 node->m_pLeft = NULL;
 node->m_pRight = NULL;

 return node;
}
BSTreeNode* createTree()
{
 BSTreeNode* root = createNode(10);

 root->m_pLeft = createNode(6);
 root->m_pRight = createNode(14);

 root->m_pLeft->m_pLeft = createNode(4);
 root->m_pLeft->m_pRight = createNode(8);

 root->m_pRight->m_pLeft = createNode(12);
 root->m_pRight->m_pRight = createNode(16);

 return root;

}

//中序遍历
void middleTraverse(BSTreeNode* root, vector<BSTreeNode* > &vec)
{
 if(root == NULL)
  return;
 middleTraverse(root->m_pLeft,vec);
 vec.push_back(root);
 middleTraverse(root->m_pRight,vec);
}


int main()
{
 BSTreeNode* root = createTree();
 vector<BSTreeNode*> vec;
 middleTraverse(root, vec);

 BSTreeNode* pHead = *(vec.begin());
 
 for(vector<BSTreeNode*>::iterator iter = vec.begin(); iter != vec.end(); ++iter)
 {
  if(iter == vec.end() -1)
   (*iter)->m_pRight = NULL;
  else
   (*iter)->m_pRight = *(iter+1);
  if(iter == vec.begin())
   (*iter)->m_pLeft = NULL;
  else
   (*iter)->m_pLeft = *(iter-1);
 }

 for(BSTreeNode* temp = pHead; temp != NULL; temp = temp->m_pRight)
  cout<<temp->m_nValue<<" ";

 cout<<endl;

 return 0;
}