1
         / \
        3   2
       / \   \  
      5   3   9 

Output: [1, 3, 9]

BFS的题目，我的做法：

    def largestValues(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        res = []
        self.helper(root,res,0)
        return res
    
    def helper(self,root,res,num):
        if not root:
            return None
        if len(res) < num+1:
            res.append(root.val)
        else:
            res[num] = max(res[num],root.val)
        self.helper(root.left,res,num+1)
        self.helper(root.right,res,num+1)```


别人的做法：
利用python中数组可以比较大小，运用map函数直接得出

```class Solution(object):
    def largestValues(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if not root:
            return []
        res = []
        left = self.largestValues(root.left)
        right = self.largestValues(root.right)
        return [root.val]+map(max,left,right)```