[leetcode]Binary Tree Zigzag Level Order Traversal

Binary Tree Zigzag Level Order Traversal

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]
]

算法思路：

BFS，没什么好讲的，用一个boolean变量标记是正序还是逆序，逆序就用头插法，正序用尾插法。

代码如下：

 1 public class Solution {
 2      List<List<Integer>> res = new ArrayList<List<Integer>>();
 3         public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
 4             if(root == null) return res;
 5             Queue<TreeNode> q = new LinkedList<TreeNode>();
 6             Queue<TreeNode> copy = new LinkedList<TreeNode>();
 7             q.offer(root);
 8             res.add(new ArrayList<Integer>(Arrays.asList(root.val)));
 9             boolean tag = false;
10             while(!q.isEmpty()){
11                 TreeNode tem = q.poll();
12                 if(tem.left != null) copy.offer(tem.left);
13                 if(tem.right != null) copy.offer(tem.right);
14                 if(q.isEmpty() && !copy.isEmpty()){
15                     List<Integer> list = new LinkedList<Integer>();
16                     if(tag){
17                         while(!copy.isEmpty()){
18                             q.add(copy.peek());
19                             list.add(copy.poll().val);
20                         }
21                     }else{
22                         while(!copy.isEmpty()){
23                             q.add(copy.peek());
24                             list.add(0, copy.poll().val);
25                         }
26                     }
27                     res.add(list);
28                     tag = !tag;
29                 }
30             }
31             return res;
32         }
33 }

我用了两个queue来进行层次的区分，网上很多人很巧妙的用了一个null来表示一层的结束。请参考这里。

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