Leetcode 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,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

return its zigzag level order traversal as:

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

Solution 1: iterative BFS use stack

 public class Solution {  
   public List<List<Integer>> zigzagLevelOrder(TreeNode root) {  
     Deque<TreeNode> s1=new LinkedList<>();  
     Deque<TreeNode> s2=new LinkedList<>();  
     List<List<Integer>> res=new ArrayList<>();  
     if (root==null) return res;  
     s1.push(root);  
     while (!s1.isEmpty()) {  
       List<Integer> one=new ArrayList<>();  
       while (!s1.isEmpty()) {  
         TreeNode x=s1.pop();  
         one.add(x.val);  
         if (x.left!=null) s2.push(x.left);  
         if (x.right!=null) s2.push(x.right);  
       }  
       res.add(one);  
       if (s2.isEmpty()) break;  
       List<Integer> two=new ArrayList<>();  
       while (!s2.isEmpty()) {  
         TreeNode x=s2.pop();  
         two.add(x.val);  
         if (x.right!=null) s1.push(x.right);  
         if (x.left!=null) s1.push(x.left);  
       }  
       res.add(two);  
     }  
     return res;  
   }  
 }  

Solution 2: recursive using DFS

 public class Solution {  
   public List<List<Integer>> zigzagLevelOrder(TreeNode root) {  
     List<List<Integer>> res=new ArrayList<>();  
     dfs(root,0,res);  
     return res;  
   }  
   private void dfs(TreeNode x, int d, List<List<Integer>> res) {  
     if (x==null) return;  
     if (d==res.size()) res.add(new LinkedList<Integer>());  
     if (d%2==0) res.get(d).add(x.val);  
     else res.get(d).add(0,x.val);  
     dfs(x.left,d+1,res);  
     dfs(x.right,d+1,res);  
   }  
 }