leetcode第102题二叉树的层序遍历--回溯算法（DFS）

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    deepth=0
    def levelOrder(self, root):
        ans=[]
        def DFS(root):
            if root:
                if len(ans)<=self.deepth:
                    ans.append([root.val])
                else:
                    ans[self.deepth].append(root.val)
                self.deepth+=1
                DFS(root.left)
                DFS(root.right)
                self.deepth-=1
        DFS(root)
        return ans

还行 一次过