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)。