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, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def delNodes(self, root: TreeNode, to_delete: List[int]) -> List[TreeNode]:
        ret = []
        def dfs(node, parent):
            if node:
                if node.val in to_delete:
                    node.left=dfs(node.left, False)
                    node.right=dfs(node.right, False)
                else:
                    if not parent:
                        ret.append(node)
                    node.left=dfs(node.left, True)
                    node.right=dfs(node.right, True)
                    return node
        dfs(root, False)
        return ret
# 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 delNodes(self, root: TreeNode, to_delete: List[int]) -> List[TreeNode]:
        ret = []
        def dfs(node, parent):
            if node:
                if node.val in to_delete:
                    node.left=dfs(node.left, False)
                    node.right=dfs(node.right, False)
                else:
                    if not parent:
                        ret.append(node)
                    node.left=dfs(node.left, True)
                    node.right=dfs(node.right, True)
                    return node
        dfs(root, False)
        return ret