Binary Tree Zigzag Level Order Traversal 二叉树按层遍历，zigzag输出@LeetCode

又是一道Level Order Traversal的变型题，可见能熟练写出Level Order Traversal是很必须的！

package Level4;

import java.util.ArrayList;
import java.util.Collections;
import java.util.LinkedList;
import java.util.Queue;

import Utility.TreeNode;

/**
 * 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]
]
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}".
 */
public class S103 {

	public static void main(String[] args) {

	}

	public ArrayList<ArrayList<Integer>> zigzagLevelOrder(TreeNode root) {
		ArrayList<ArrayList<Integer>> ret = new ArrayList<ArrayList<Integer>>();
		if(root == null){
			return ret;
		}
		
		Queue<TreeNode> queue = new LinkedList<TreeNode>();
		ArrayList<Integer> al = new ArrayList<Integer>();
		queue.add(root);
		int currentLevel = 1;
		int nextLevel = 0;
		boolean left = true;
		
		while(!queue.isEmpty()){
			TreeNode cur = queue.remove();
			currentLevel--;
			al.add(cur.val);
			if(cur.left != null){
				queue.add(cur.left);
				nextLevel++;
			}
			if(cur.right!=null){
				queue.add(cur.right);
				nextLevel++;
			}
			
			if(currentLevel == 0){
				if(!left){			// 当自右往左时，要翻转al
					Collections.reverse(al);
				}
				left = !left;
				ret.add(al);
				al = new ArrayList<Integer>();
				currentLevel = nextLevel;
				nextLevel = 0;
			}
		}
		return ret;
	}
}

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> ret = new ArrayList<List<Integer>>();
        
        if(root == null) {
            return ret;
        }
        
        LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
        queue.add(root);
        
        int curLevelCnt = 1;
        int nextLevelCnt = 0;
        List<Integer> list = new ArrayList<Integer>();
        boolean order = true;
        
        while(!queue.isEmpty()) {
            TreeNode cur = queue.remove();
            list.add(cur.val);
            curLevelCnt--;
            
            if(cur.left != null) {
                queue.add(cur.left);
                nextLevelCnt++;
            }
            if(cur.right != null) {
                queue.add(cur.right);
                nextLevelCnt++;
            }
            
            if(curLevelCnt == 0) {
                curLevelCnt = nextLevelCnt;
                nextLevelCnt = 0;
                if(!order) {
                    Collections.reverse(list);
                }
                order = !order;
                ret.add(new ArrayList(list));
                list.clear();
            }
        }
        
        return ret;
    }
}