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