leetcode_103_Binary Tree Zigzag Level Order Traversal

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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：

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