/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
 /*
dfs遍历二叉树，从左至右。每个元素都有个deep（深度），只需将root.val放入result.get(deep)数组中即可。
时间复杂度O（n），击败97%用户。
 */
class Solution {
    public static List<List<Integer>> result;
    public static void dfs(TreeNode root,int deep){
        if(root.left == null && root.right == null)return;
        if(root.left!=null){
            if(deep+1 > (result.size())-1)result.add(new LinkedList<Integer>());
            if(deep%2!=0)result.get(deep+1).add(root.left.val);
            else result.get(deep+1).add(0,root.left.val);
            dfs(root.left,deep+1);
        }
        if(root.right!=null){
            if(deep+1 > (result.size())-1)result.add(new LinkedList<Integer>());
            if(deep%2!=0)result.get(deep+1).add(root.right.val);
            else result.get(deep+1).add(0,root.right.val);
            dfs(root.right,deep+1);
        }
    }
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        result = new ArrayList<List<Integer>>();
        if(root == null)return result;
        result.add(new LinkedList<Integer>());
        result.get(0).add(root.val);
        dfs(root,0);
        return result;
    }
}