leetcode第46題（binary-tree-zigzag-level-order-traversal）

題目：

Given a binary tree, return the zigzag level ordertraversal 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：用兩個隊列或者兩個堆實作。或者是用一個隊列或者堆，奇數行的時候實作翻轉即可。

代碼：

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
import java.util.ArrayList;
import java.util.Collections;
import java.util.LinkedList;
import java.util.Queue;
public class Solution {
    public ArrayList<ArrayList<Integer>> zigzagLevelOrder(TreeNode root) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<>();
        Queue<TreeNode> queue = new LinkedList<>();
        boolean left = true;
        if (root != null) {
            queue.add(root);
        }
        while (!queue.isEmpty()) {
            int n = queue.size();
            ArrayList<Integer> temp = new ArrayList<>();
            for(int i=n-1;i>=0;i--) {
                TreeNode node = queue.poll();
                if (node.left != null) {
                    queue.add(node.left);
                }
                if (node.right != null) {
                    queue.add(node.right);
                }
                temp.add(node.val);
            }
            if (result.size() % 2 != 0) {
                Collections.reverse(temp);
            }
            result.add(temp);
        }
        return result;
    }
}