We are given the head node root of a binary tree, where additionally every node's value is either a 0 or a 1.
Return the same tree where every subtree (of the given tree) not containing a 1 has been removed.
(Recall that the subtree of a node X is X, plus every node that is a descendant of X.)
Example 1:
Input: [1,null,0,0,1]
Output: [1,null,0,null,1]
Explanation: 
Only the red nodes satisfy the property "every subtree not containing a 1".
The diagram on the right represents the answer.
Example 2:
Input: [1,0,1,0,0,0,1]
Output: [1,null,1,null,1]
Example 3:
Input: [1,1,0,1,1,0,1,0]
Output: [1,1,0,1,1,null,1]
Note: 
The binary tree will have at most 200 nodes.
The value of each node will only be 0 or 1.
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def pruneTree(self, root: TreeNode) -> TreeNode:
        if root == None:
            return None
        left_sub = self.pruneTree(root.left)
        right_sub = self.pruneTree(root.right)
        if root.val  == 0 and left_sub == None and right_sub == None:
            return None
        root.left  = left_sub
        root.right = right_sub
        return root
        

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def pruneTree(self, root: TreeNode) -> TreeNode:
        if root == None:
            return None
        left_sub = self.pruneTree(root.left)
        right_sub = self.pruneTree(root.right)
        if root.val  == 0 and left_sub == None and right_sub == None:
            return None
        root.left  = left_sub
        root.right = right_sub
        return root
        
