原题出处：https://leetcode.cn/leetbook/read/top-interview-questions-medium/xvle7s/

解法一：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<>();
        if (root == null) {
            return result;
        }
        Queue<TreeNode> q = new LinkedList();
        q.add(root);
        boolean isLeft = true;
        while (!q.isEmpty()) {
            List<Integer> list = new ArrayList();
            int count = q.size();
            for (int i = 0 ; i < count; i++) {
                TreeNode treeNode = q.poll();
                if (isLeft) {
                    list.add(treeNode.val);
                } else {
                    list.add(0,treeNode.val);
                }
                if (treeNode.left != null) {
                    q.add(treeNode.left);
                }
                if (treeNode.right != null) {
                    q.add(treeNode.right);
                }
            }
            isLeft = !isLeft;
            result.add(list);
        }
        return result;
    }
}

思路：BFS,广度优先，我们使用一个变量来保存向左还是向右进行遍历，每次遍历完之后进行翻转即可。