剑指 Offer 32 - III. 从上到下打印二叉树 III

请实现一个函数按照之字形顺序打印二叉树，即第一行按照从左到右的顺序打印，第二层按照从右到左的顺序打印，第三行再按照从左到右的顺序打印，其他行以此类推。

例如:
给定二叉树: [3,9,20,null,null,15,7],

返回其层次遍历结果：

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

1、Collections.reverse()方法翻转

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        Deque<TreeNode> queue = new LinkedList<>();
        List<List<Integer>> res = new ArrayList<>();
        if (root == null) {
            return res;
        } else {
            queue.offer(root);
        }
        while (!queue.isEmpty()) {
            List<Integer> list = new ArrayList<>();
            for (int i = queue.size(); i > 0; 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);
                }
            }
            res.add(list);
        }
        for (int i = 1; i < res.size(); i+=2) {
            Collections.reverse(res.get(i));
        }
        return res;
    }
}

2、LinkedList双端列表

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        Deque<TreeNode> queue = new LinkedList<>();
        List<List<Integer>> res = new ArrayList<>();
        if (root == null) {
            return res;
        } else {
            queue.offer(root);
        }
        while (!queue.isEmpty()) {
            LinkedList<Integer> list = new LinkedList<>();
            for (int i = queue.size(); i > 0; i--) {
                TreeNode node = queue.poll();
                if (res.size()%2 == 1) {
                    list.addFirst(node.val);
                } else {
                    list.addLast(node.val);
                }
                if (node.left != null) {
                    queue.offer(node.left);
                }
                if (node.right != null) {
                    queue.offer(node.right);
                }
            }
            res.add(list);
        }
        return res;
    }
}