Given the root of a binary tree, each node in the tree has a distinct value.
After deleting all nodes with a value in to_delete, we are left with a forest (a disjoint union of trees).
Return the roots of the trees in the remaining forest. You may return the result in any order.
Example 1:

Input: root = [1,2,3,4,5,6,7], to_delete = [3,5] Output: [[1,2,null,4],[6],[7]]
Constraints:
- The number of nodes in the given tree is at most
1000. - Each node has a distinct value between
1and1000. to_delete.length <= 1000to_deletecontains distinct values between1and1000.
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def delNodes(self, root: TreeNode, to_delete: List[int]) -> List[TreeNode]:
forest = []
to_delete = set(to_delete)
def buttom_up(root):
if root:
left, right = dfs(root.left), dfs(root.right)
root.left, root.right = left, right
if root.val in to_delete:
if left: forest.append(left)
if right: forest.append(right)
else:
return root
dfs(root)
return ([] if root.val in to_delete else [root]) + forest
#########################################################
to_delete_set = set(to_delete)
res = []
def top_down(root, is_root):
if not root:
return None
root_deleted = root.val in to_delete_set
if is_root and not root_deleted:
res.append(root)
root.left = helper(root.left, root_deleted)
root.right = helper(root.right, root_deleted)
return None if root_deleted else root
top_down(root, True)
return res