Binary Tree Zigzag Level Order Traversal

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]
]

[Thoughts]:
变形BFS，creat一个boolean 变量tracking order。

public class Solution {
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        if(root==null) return res;
        
        ArrayList<TreeNode> list = new ArrayList<TreeNode>();
        list.add(root);
        boolean order = true; //add left first, then right;
        
        while(!list.isEmpty()){
            ArrayList<TreeNode> temp = new ArrayList<TreeNode>();
            List<Integer> intList = new ArrayList<Integer>();
            for(int i=list.size()-1; i>=0; i--){//Note: 逆序
                TreeNode node = list.get(i);
                intList.add(node.val);
                if(order){
                    if(node.left!=null)
                        temp.add(node.left);
                    if(node.right!=null)
                        temp.add(node.right);
                }else{
                    if(node.right!=null)
                        temp.add(node.right);
                    if(node.left!=null)
                        temp.add(node.left);
                }
            }
            res.add(intList);
            list = temp;
            order = !order;//Note: order每次要取反
        }
        
        return res;   
    }
}