leetcode_103_Binary Tree Zigzag Level Order Traversal

描述：

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.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / \
 2   3
    /
   4
    \
     5

The above binary tree is serialized as  "{1,2,3,#,#,4,#,#,5}".

思路：

具体的思路和层序遍历一致，只是需要将偶数层的结点值颠倒一下即可，时间略长，可见本方法并非好的方法。思路详见层序遍历： http://blog.csdn.net/mnmlist/article/details/44490975

代码：

public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
		List<List<Integer>> list = new ArrayList<List<Integer>>();// 存储结果
		if (root == null)
			return list;
		Queue<TreeNode> q1 = new LinkedList<TreeNode>();// 交替存储相邻两层的结点
		Queue<TreeNode> q2 = new LinkedList<TreeNode>();
		Queue<TreeNode> temp = null;
		List<Integer> subList = null;// 存储一层的结点的值
		q1.add(root);
		TreeNode top = null;
		int flag=0;
		while (!q1.isEmpty()) {
			subList = new ArrayList<Integer>();
			while (!q1.isEmpty())// 循环遍历一层结点并将下一层结点存储到队列中
			{
				top = q1.peek();
				q1.poll();
				if (top.left != null)
					q2.add(top.left);
				if (top.right != null)
					q2.add(top.right);
				subList.add(top.val);
			}
			flag++;
			if((flag&1)==0)
				Collections.reverse(subList);
			list.add(subList);
			temp = q2;// 交换两个队列的值，使q1一直指向要遍历的那一层
			q2 = q1;
			q1 = temp;
		}
		return list;
	}

结果：