Leetcode 107 Binary Tree Level Order Traversal II

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

return its bottom-up level order traversal as:

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

Solution 1: iterative BFS

 public class Solution {  
   public List<List<Integer>> levelOrderBottom(TreeNode root) {  
     List<List<Integer>> res=new LinkedList<>();  
     Queue<TreeNode> q1=new LinkedList<>();  
     Queue<TreeNode> q2=new LinkedList<>();  
     if (root==null) return res;  
     q1.offer(root);  
     while (!q1.isEmpty()) {  
       List<Integer> one=new ArrayList<>();  
       while (!q1.isEmpty()) {  
         TreeNode x=q1.poll();  
         one.add(x.val);  
         if (x.left!=null) q2.offer(x.left);  
         if (x.right!=null) q2.offer(x.right);  
       }  
       res.add(0,one);  
       Queue<TreeNode> temp=q1;  
       q1=q2;  
       q2=temp;  
     }  
     return res;  
   }  
 }  

Solution 2: Recursive DFS, compare to BFS version, the insertion of new line cost is high.

 public class Solution {  
   public List<List<Integer>> levelOrderBottom(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(0,new ArrayList<Integer>());//cost high for ArrayList  
     res.get(res.size()-d-1).add(x.val);//cost low, happen more frequently  
     dfs(x.left,d+1,res);  
     dfs(x.right,d+1,res);  
   }  
 }