二叉树中序遍历下一个节点

#include<stdio.h>
#include<string.h>
#include <pthread.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <stack>
#include <stdlib.h>
#include <sstream>
using namespace std;

struct TreeLinkNode {
    int val;
    struct TreeLinkNode *left;
    struct TreeLinkNode *right;
    struct TreeLinkNode *next;
    TreeLinkNode(int x) :val(x), left(NULL), right(NULL), next(NULL) {

    }
};

class Solution {
    //分为两大类：1，有右子树的，那么下个节点就是右子树最左边啊的点
    //2.没有右子树的a)是父节点的左孩子，那么父节点就是下一个节点b)时父节点的右孩子，找他的父节点的父节点的父节点，。。。。
    //直到当前节点时其父节点的左孩子的位置如果没有，那他就是尾节点
public:
    TreeLinkNode* GetNext(TreeLinkNode* pNode)
    {
        if(pNode==NULL)
            return NULL;
        if(pNode->right!=NULL){
            pNode=pNode->right;
            while(pNode->left!=NULL)
                pNode=pNode->left;
            return pNode;
        }
        while(pNode->next!=NULL){
            TreeLinkNode * proot=pNode->next;
            if(proot->left==pNode)
                return proot;
            pNode=pNode->next;
        }
        return NULL;
    }
};

int main()
{
    TreeLinkNode*t1=new TreeLinkNode(8);
    TreeLinkNode*t2=new TreeLinkNode(6);
    TreeLinkNode*t3=new TreeLinkNode(10);
    TreeLinkNode*t4=new TreeLinkNode(5);
    TreeLinkNode*t5=new TreeLinkNode(7);
    TreeLinkNode*t6=new TreeLinkNode(9);
    TreeLinkNode*t7=new TreeLinkNode(11);
    t1->left=t2;
    t1->right=t3;
    t2->left = t4;
    t2->right = t5;
    t3->left = t6;
    t3->right = t7;
    t2->next = t1;
    t3->next = t1;
    t4->next = t2;
    t5->next = t2;
    t6->next = t3;
    t7->next = t3;
    Solution s;
    if(s.GetNext(t1)==NULL)
        cout<<"NO next point "<<endl;
    else
        cout<<"s.GetNext(t1): "<<s.GetNext(t1)->val<<endl;
    return 0;

}