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,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its bottom-up level order traversal as:

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

一刷
題解：
BFS， add新的list的時(shí)候應(yīng)該insert到頭部，那么層較深，就會(huì)創(chuàng)建新的頭部并且插在頭部。list的size得到樹的高度h，然后用h-level得到應(yīng)該插入哪個(gè)list.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> res = new LinkedList<>();
        traverse(root, 0, res);
        return res;
    }
    
    private void traverse(TreeNode root, int level, List<List<Integer>> res){
        if(root==null) return;
        if(level>=res.size()) res.add(0, new ArrayList<>());
        res.get(res.size()-1-level).add(root.val);
        traverse(root.left, level+1, res);
        traverse(root.right, level+1, res);
    }
}