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]]

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:
很简单的一个，比Binary Tree Level Order Traversal多了一个方向标志(Zigzag的方向性)，

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {        List<List<Integer>> list = new LinkedList<List<Integer>>();//返回结果        Queue<TreeNode> queue = new LinkedList<TreeNode>();//队列，使用BFS遍历        int level = 0;//记录某一层进队的元素数        int count = 0;//计数        if(root == null){            return list;        }        queue.offer(root);        level++;        TreeNode p = null;        boolean flag = true;//标志Zigzap的方向        while(queue.size() > 0){            int nextLevel = 0;//下一层进队的元素数            List<Integer> tmp = new ArrayList<Integer>();//中间存储每一行的结果            while(count < level){                p = queue.poll();//出队                if(p.left != null){                    queue.offer(p.left);                    nextLevel++;                }                if(p.right != null){                    queue.offer(p.right);                    nextLevel++;                }                if(flag){                    tmp.add(p.val);                }else{                    tmp.add(0,p.val);                }                count ++;            }            list.add(tmp);            count = 0;            level = nextLevel;            flag = !flag;        }        return list;    }}

Runtime: 227 ms