輸出8 6 10 5 7 9 11。

解題思路：利用佇列先進先出(FIFO)的性質，取出隊首元素，輸出隊首節點元素，將節點的左右子節點加入佇列，取出隊首元素，重複上述操作，直到取出佇列中所有元素。

#include <queue>
#include <iostream>
#include <string.h>
using namespace std;
struct BSTreeNode          //a node in the binary search tree (BST)
{
  int m_nValue;            //value of node
  BSTreeNode *m_pLeft;     // left child of node
  BSTreeNode *m_pRight;    // right child of node
};
queue<BSTreeNode*>P;
void BuildTree(BSTreeNode *&Node,int n)   //建立二叉樹
{
	if(Node == NULL)
	{
		BSTreeNode *NewNode = new BSTreeNode();
		NewNode->m_nValue = n;
		NewNode->m_pLeft = NULL;
		NewNode->m_pRight = NULL;
		Node = NewNode;
	}
	else if(Node->m_nValue < n) 
	{
		BuildTree(Node->m_pRight,n);
	}
	else 
	{
		BuildTree(Node->m_pLeft,n);
	}
}
void OutPut()
{
	if(P.front() == NULL)
		return ;
	BSTreeNode *Node = P.front();
	cout<<P.front()->m_nValue<<" ";
	P.push(P.front()->m_pLeft);
	P.push(P.front()->m_pRight);
	P.pop();
	OutPut();
}
int main()
{
	BSTreeNode *Node = NULL;
	BuildTree(Node,8);
	BuildTree(Node,6);
	BuildTree(Node,10);
	BuildTree(Node,5);
	BuildTree(Node,7);
	BuildTree(Node,9);
	BuildTree(Node,11);
	P.push(Node);
	OutPut();
	cout<<endl;
	return 0;
}