Binary Tree Zigzag Level Order Traversal.txt

problem:
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]
]
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

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

//两个栈(curLevel, nextLevel)来模拟，并用一个变量记录是l-to-r或r-to-l
vector<vector<int> > zigzagLevelOrder(TreeNode *root)
{
	vector<vector<int> > ret;
	if(root == NULL)
		return ret;
	stack<TreeNode*> curLevel, nextLevel;
	bool isLToR = true;
	curLevel.push(root);
	vector<int> tmp;
	while(!curLevel.empty())
	{
		TreeNode *curNode = curLevel.top();
		curLevel.pop();
		if(curNode)
		{
			tmp.push_back(curNode->val);
			if(isLToR)
			{
				if(curNode->left)
					nextLevel.push(curNode->left);
				if(curNode->right)
					nextLevel.push(curNode->right);
			}
			else
			{
				if(curNode->right)
					nextLevel.push(curNode->right);
				if(curNode->left)
					nextLevel.push(curNode->left);
			}
		}
		if(curLevel.empty())
		{
			ret.push_back(tmp);
			tmp.clear();
			isLToR = !isLToR;
			swap(curLevel, nextLevel);
		}
	}
	return ret;
}