Binary Tree Zigzag Level Order Traversal

description:
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).

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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]
]

解题思路：
实用BFS的方式进行搜索处理，将所得到的结果进行按照奇偶层判断是否需要翻转处理。

/** * Definition of TreeNode: * public class TreeNode { *     public int val; *     public TreeNode left, right; *     public TreeNode(int val) { *         this.val = val; *         this.left = this.right = null; *     } * } */public class Solution {    /**     * @param root: The root of binary tree.     * @return: A list of lists of integer include      *          the zigzag level order traversal of its nodes' values      */    public ArrayList<ArrayList<Integer>> zigzagLevelOrder(TreeNode root) {        // write your code here        ArrayList<ArrayList<Integer>> result = new ArrayList<>();        if (root == null) {            return result;        }        Queue<TreeNode> queue = new LinkedList<>();        queue.offer(root);        int index = 0;        while (!queue.isEmpty()) {            ArrayList<Integer> list = new ArrayList<>();            int len = queue.size();            for (int i = 0; i < len; i++) {                TreeNode node = queue.poll();                list.add(node.val);                if (node.left != null) {                    queue.offer(node.left);                }                if (node.right != null) {                    queue.offer(node.right);                }            }            if (index % 2 != 0) {                Collections.reverse(list);            }            index++;            result.add(list);        }        return result;    }}