【LeetCode】Binary Tree Zigzag Level Order Traversal

           Binary Tree Zigzag Level Order Traversal 
           Total Accepted: 3689 Total Submissions: 14306 My Submissions
           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},
           
           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.
           这个题是螺旋打印，其实和Binary Tree Level Order Traversal，Binary Tree Level Order Traversal II思路一样。
           每次放在队列中的时候，判断一下，当前是哪一层。
           假设层数为K，并且从1开始，那么在k % 2 == 1的时候，先放左子树，后放右子树，全部放入tempList以后，最后逆序放在队列中。
           如果k % 2 == 0，先放右子树，后放左子树，全部放入tempList以后，最后逆序放在队列中。
           这个过程如下：
           假设我们有一棵这样的树。分析过程如下。

           
           BFS
          Java AC

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<ArrayList<Integer>> zigzagLevelOrder(TreeNode root) {
        ArrayList<ArrayList<Integer>> list = new ArrayList<ArrayList<Integer>>();
        if(root == null){
            return list;
        }
        bfs(root,list);
        return list;
    }
    public void bfs(TreeNode root,ArrayList<ArrayList<Integer>> list){
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
		queue.offer(root);
		int k = 0;
		while (!queue.isEmpty()) {
			k++;
			int size = queue.size();
			ArrayList<Integer> tempList = new ArrayList<Integer>();
			ArrayList<TreeNode> nodeList = new ArrayList<TreeNode>();
			if (k % 2 == 1) {
				for (int i = 0; i < size; i++) {
					TreeNode point = queue.peek();
					queue.poll();
					tempList.add(point.val);
					if (point.left != null) {
						nodeList.add(point.left);
					}
					if (point.right != null) {
						nodeList.add(point.right);
					}
				}
			} else {
				for (int i = 0; i < size; i++) {
					TreeNode point = queue.peek();
					queue.poll();
					tempList.add(point.val);
					if (point.right != null) {
						nodeList.add(point.right);
					}
					if (point.left != null) {
						nodeList.add(point.left);
					}
				}
			}
			int listSize = nodeList.size();
			for (int i = listSize-1; i >= 0; i--) {
				queue.offer(nodeList.get(i));
			}
			list.add(tempList);
		}
	}
}