对于一颗完全二叉树，要求给所有节点加上一个pNext指针，指向同一层的相邻节点-----层序遍历的应用题

题目：对于一颗完全二叉树，要求给所有节点加上一个pNext指针，指向同一层的相邻节点；如果当前节点已经是该层的最后一个节点，则将pNext指针指向NULL；给出程序实现，并分析时间复杂度和空间复杂度。

#include "stdafx.h"
#include <iostream>
#include <fstream>
#include <vector>

using namespace std;

struct TreeNode
{
int m_nValue;
TreeNode *m_pLeft;
TreeNode *m_pRight;
TreeNode *pNext;
};

//假定所创建的二叉树如下图所示
/*
1
/     \
2       3
/ \      / \
4   5    6   7
/ \  / \  / \
8   9 10 11 12 13
*/
void CreateBitree(TreeNode *&pNode, fstream &fin)
{
int dat;
fin>>dat;
if(dat == 0)
{
pNode = NULL;
}
else
{
pNode = new TreeNode();
pNode->m_nValue = dat;
pNode->m_pLeft = NULL;
pNode->m_pRight = NULL;
pNode->pNext = NULL;
CreateBitree(pNode->m_pLeft, fin);
CreateBitree(pNode->m_pRight, fin);
}
}

//完全二叉树指向同一层的相邻结点
void Solution(TreeNode *pHead)
{
if (NULL == pHead)
{
return;
}
vector<TreeNode*> vec;
vec.push_back(pHead);
TreeNode *pre = NULL;
TreeNode *pNode = NULL;
int cur = 0;
int last = 0;
while(cur < vec.size())
{
last = vec.size();
while (cur < last)
{
if (NULL == pre)
{
pre = vec[cur];
}
else
{
pre->pNext = vec[cur];
}
if (NULL != vec[cur]->m_pLeft)
{
vec.push_back(vec[cur]->m_pLeft);
}
if (NULL != vec[cur]->m_pRight)
{
vec.push_back(vec[cur]->m_pRight);
}
pre = vec[cur];
cur++;
}
pre->pNext = NULL;
pre = NULL;
}
}

int _tmain(int argc, _TCHAR* argv[])
{
fstream fin("tree.txt");
TreeNode *pHead = NULL;
TreeNode *pNode = NULL;
CreateBitree(pHead, fin);
Solution(pHead);

while (NULL != pHead)
{
cout<<pHead->m_nValue<<"  ";
pNode = pHead->pNext;
while (NULL != pNode)
{
cout<<pNode->m_nValue<<"  ";
pNode = pNode->pNext;
}
cout<<endl;
pHead = pHead->m_pLeft;
}

cout<<endl;
return 0;
}

答：时间复杂度为O(n),空间复杂度为O(n)。