基于中序遍历找到一个结点的后继结点

2.当p没有右子树时。有以下两种情况：

/*
找到中序遍历下一个结点的后继结点
by Rowandjj
2014/8/19
*/
#include<iostream>
using namespace std;
typedef struct _NODE_
{
	int data;
	struct _NODE_ *left;
	struct _NODE_ *right;
	struct _NODE_ *parent;
}Node,*pNode,*pTree;
//找到中序遍历时，p的后继结点
pNode after(pNode p)
{
	if(p == NULL)
	{
		return NULL;
	}
	if(p->right != NULL)//存在右子树
	{//找到右子树中的最左结点
		pNode pTemp = p->right;
		while(pTemp->left)
		{
			pTemp = pTemp->left;
		}
		return pTemp;
	}else//不存在右子树
	{//那么找到该结点的父结点，若当前结点是父结点的左孩子那么父结点即为所求
		//否则继续向上寻找
		pNode pParent = p->parent;
		while(pParent && pParent->right == p)
		{	
			p = pParent;
			pParent = p->parent;
		}
		return pParent;
	}
}
//构建
void create(pTree *pRoot,pNode pParent)
{
	int data;
	cin>>data;
	if(data == -1)
	{
		return;
	}
	*pRoot = (pNode)malloc(sizeof(Node));
	if(*pRoot == NULL)
	{
		exit(-1);
	}
	(*pRoot)->data = data;
	(*pRoot)->left = NULL;
	(*pRoot)->right = NULL;
	(*pRoot)->parent = pParent;
	create(&(*pRoot)->left,*pRoot);
	create(&(*pRoot)->right,*pRoot);
}
//中序遍历
void display(pTree pRoot)
{
	if(pRoot == NULL)
	{
		return;
	}
	if(pRoot->left != NULL)
	{
		display(pRoot->left);
	}
	cout<<pRoot->data<<" ";
	if(pRoot->right != NULL)
	{
		display(pRoot->right);
	}
}
int main()
{
	pTree pTree = NULL;
	create(&pTree,NULL);
// 	pNode p1 = pTree->left->right;
// 	cout<<p1->data<<endl;
// 	cout<<p1->parent->data<<endl;
//	cout<<after(pTree)->data;		
	//cout<<after(pTree->left->right)->data<<endl;
	cout<<after(pTree->left)->data<<endl;
	display(pTree);
	return 0;
}