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]
]

这道题主要用BFS解决，在BFS得到结果后需要考虑是否要逆序一下。逆序可以考虑用栈，我没有用栈，直接逆序

public class Solution {
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();			//保存最后的结果
        if(null == root)
        	return result;
        boolean needReverse = false;												//是否要对列表中的内容进行逆序
        Queue<TreeNode> queue = new LinkedList<TreeNode>();						//BFS要使用到的队列
        queue.add(root);														//根节点入队
        while(!queue.isEmpty()){												//队列不为空
        	List<Integer> oneLevel = new ArrayList<Integer>();					//保存在同一层及节点的值
        	Queue<TreeNode> temp = new LinkedList<TreeNode>();					//存放下一层的节点
        	while(!queue.isEmpty()){
        		TreeNode headOfQueue = queue.remove();							//对头元素出队
        		oneLevel.add(headOfQueue.val);
        		if(headOfQueue.left != null)
        			temp.add(headOfQueue.left);									//左子树不为空，入队
        		if(headOfQueue.right != null)
        			temp.add(headOfQueue.right);								//右子树不为空，入队
        	}//while
        	if(needReverse)
        		reverseList(oneLevel);
        	result.add(oneLevel);
        	needReverse = !needReverse;
        	queue = temp;
        }
        return result;
    }

    /**
     * list列表中内容逆序
     * @param list
     */
    private void reverseList(List<Integer> list){
    	for(int i = 0, j = list.size() - 1; i < j; i++, j--){
    		int temp = list.get(i);
    		list.set(i, list.get(j));
    		list.set(j, temp);
    	}
    }
}