leetcode------Binary Tree Zigzag Level Order Traversal

标题：	Binary Tree Zigzag Level Order Traversal
通过率：	26.5%
难度：	中等

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.

本题就是按层次遍历二叉树，每次遍历时进行queue的翻转即可。

具体看代码:

 1 /**

 2  * Definition for binary tree

 3  * public class TreeNode {

 4  *     int val;

 5  *     TreeNode left;

 6  *     TreeNode right;

 7  *     TreeNode(int x) { val = x; }

 8  * }

 9  */

10 public class Solution {

11     public List<List<Integer>> zigzagLevelOrder(TreeNode root) {

12        List<List<Integer>> res = new ArrayList<List<Integer>>();

13         if (root == null) {

14             return res;

15         }

16         List<Integer> tmp = new ArrayList<Integer>();

17         Queue<TreeNode> queue = new LinkedList<TreeNode>();

18         queue.offer(root);

19         int num;

20         boolean reverse = false;

21         while (!queue.isEmpty()) {

22             num = queue.size();

23             tmp.clear();

24             for (int i = 0; i < num; i++) {

25                 TreeNode node = queue.poll();

26                 tmp.add(node.val);

27                 if (node.left != null)

28                     queue.offer(node.left);

29                 if (node.right != null)

30                     queue.offer(node.right);

31             }

32             if (reverse) {

33                 Collections.reverse(tmp);

34                 reverse = false;

35             }

36             else

37                 reverse = true;

38             res.add(new ArrayList<Integer>(tmp));

39         }

40         return res;

41         

42     }

43 }