# Delete Nodes And Return Forest

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 `1` and `1000`.
• `to_delete.length <= 1000`
• `to_delete` contains distinct values between `1` and `1000`.
```# 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```