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]
]

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

Solution:
Use bfs to traverse level by level and reverse temp result in even rows.

/**
public class Solution {
    public ArrayList> zigzagLevelOrder(TreeNode root) {
        ArrayList> result = new ArrayList>();
        ArrayList toVisit = new ArrayList();
        if(root == null) 
            return result;
        toVisit.add(root);
        boolean flag = true;
        while(!toVisit.isEmpty()){
            ArrayList tmp = new ArrayList();
            for(TreeNode tn : toVisit)
                tmp.add(tn.val);
            result.add(tmp);
            
            // why start from the last element? 
            // write an example on paper then you will see.
            ArrayList next = new ArrayList();
            
            for(int i = toVisit.size()-1; i >= 0 ; i--){
                TreeNode tn = toVisit.get(i);
                if(flag){
                    if(tn.right != null) next.add(tn.right);
                    if(tn.left != null) next.add(tn.left);
                }
                else{
                    if(tn.left != null) next.add(tn.left);
                    if(tn.right != null) next.add(tn.right);
                }
            }
            
            // change flag alternatively
            flag = flag ? false : true;
            
            toVisit = new ArrayList(next);
        }
        return result;
    }
}