锯齿状层次化输出二叉树

/*given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).

For example:
Given binary tree {3,9,20,#,#,15,7},
	      3
		 / \
	    9  20
		  /  \
		 15   7
return its zigzag level order traversal as:
[
  [3],
  [20,9],
  [15,7]
]
*/
//思路：使用双段队列
//设置层次，如果是奇数层，从左往右
//如果偶数层，从右往左（左右指针相同）

#include 
#include 
#include 

struct TreeNode
{
	int val;
	TreeNode* left;
	TreeNode* right;
	TreeNode(int x):val(x),left(NULL),right(NULL){}
};

class Solution
{
public:
	std::vector > zigzagLevelOrder(TreeNode* root)
	{
		std::vector > vec;
		if(root==NULL)
			return vec;
		int level=1; //odd or ....
		int count;
		TreeNode* tem;
		std::deque deque;
		deque.push_back(root);
		while(!deque.empty())
		{
			count=deque.size();
			std::vector lv;
			if(level%2==0)//from right to left;
			{
				while(count--)
				{
					tem=deque.back();
					deque.pop_back();
					lv.push_back(tem->val);
					if(tem->right!=NULL)
						deque.push_front(tem->right);
					if(tem->left!=NULL)
						deque.push_front(tem->left);
				}
				vec.push_back(lv);
			}
			else //from left to right
			{
				while(count--)
				{
					tem=deque.front();
					deque.pop_front();
					lv.push_back(tem->val);
					if(tem->left!=NULL)
						deque.push_back(tem->left);
					if(tem->right!=NULL)
						deque.push_back(tem->right);
				}
				vec.push_back(lv);
			}
			++level;
		}
		return vec;
	}
};

int main(int argc,char* argv[])
{
	return 0;
}