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。

运用队列，按层遍历，每次遍历一层时，添加新指针，由于每个节点只需要进队一次出队一次，时间复杂度为O(n)，空间复杂度为O(n)，具体代码如下：

#include<iostream>
#include<vector>
#include<queue>
using namespace std;

struct BinaryTreeNode
{
	int data;
	BinaryTreeNode* pNext;
	BinaryTreeNode* lchild,*rchild;
	BinaryTreeNode(int x):data(x),pNext(NULL),lchild(NULL),rchild(NULL){}
};

void addNextPoint(BinaryTreeNode* root)
{
	if(root == NULL)return;
	queue<BinaryTreeNode*> q;
	q.push(root);
	while(!q.empty())
	{
		int levelLength = q.size();//当前层的节点个数
		BinaryTreeNode* first = NULL,*second = NULL;//每次取两个元素进行操作
		while(levelLength > 0)
		{
			first = q.front();
			q.pop();
			if(first->lchild)q.push(first->lchild);
			if(first->rchild)q.push(first->rchild);
			if(--levelLength == 0)break;
			second = q.front();//第二个元素不需要出队
			first->pNext = second;
		}
	}
}

void printLevel(BinaryTreeNode* root)//打印测试
{
	if(root)
	{
		cout << root->data << " pNext 为：";
		if(root->pNext) cout << root->pNext->data << " " << endl;
		else cout << " NULL"<<endl;
		printLevel(root->lchild);
		printLevel(root->rchild);
	}
}

int main()
{
	BinaryTreeNode* root = new BinaryTreeNode(1);
	root -> lchild = new BinaryTreeNode(2);
	root -> rchild = new BinaryTreeNode(3);
	root -> lchild -> lchild = new BinaryTreeNode(4);
	root -> lchild -> rchild = new BinaryTreeNode(5);
	root -> rchild -> lchild = new BinaryTreeNode(6);
	root -> rchild -> rchild = new BinaryTreeNode(7);
	root -> lchild -> lchild -> lchild = new BinaryTreeNode(8);
	addNextPoint(root);
	printLevel(root);
}