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 }

 

posted @ 2015-03-14 13:57  pku_smile  阅读(160)  评论(0)  编辑  收藏  举报