103. 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,null,null,15,7],

    3   / \  9  20    /  \   15   7

return its zigzag level order traversal as:

[  [3],  [20,9],  [15,7]]

跟传统102. Binary Tree Level Order Traversal相比，只需要改动向ArrayList的添加方向就能实现蛇形遍历，单数层ArrayList.add(val)，双数层ArrayList.add(0, val)。代码如下：

/** * Definition for a binary tree node. * 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>> res = new ArrayList<List<Integer>>();        if (root == null) return res;        Queue<TreeNode> queue = new LinkedList<TreeNode>();        boolean reverse = false;        queue.offer(root);        while (!queue.isEmpty()) {            int size = queue.size();            List<Integer> subLevel = new ArrayList<Integer>();            for (int i = 0; i < size; i++) {                if (queue.peek().left != null) queue.offer(queue.peek().left);                if (queue.peek().right != null) queue.offer(queue.peek().right);                if (reverse) {                    subLevel.add(0, queue.poll().val);                } else {                    subLevel.add(queue.poll().val);                }            }            reverse = !reverse;            res.add(subLevel);        }        return res;    }}