Python Solution
Intuition\nThe problem requires us to remove certain nodes from a binary tree and return the roots of the resulting forest. By traversing the tree and handling 
### Intuition\nThe problem requires us to remove certain nodes from a binary tree and return the roots of the resulting forest. By traversing the tree and handling deletions recursively, we can effectively manage the separations of the tree into multiple trees. Each time we delete a node, its children (if any) become potential new roots of the trees in the forest.\n\n### Approach\n1. **Convert `to_delete` to a Set**:\n   - This allows us to quickly check if a node needs to be deleted.\n\n2. **Define a Recursive Function (`dfs`)**:\n   - This function will traverse the tree.\n   - For each node, determine if it needs to be deleted.\n   - If a node is deleted, its children are considered potential new roots and added to the forest if they are not `None`.\n   - If a node is not deleted, continue traversing its children.\n   - The function returns `None` if the node is deleted, otherwise it returns the node itself.\n   \n3. **Handle the Initial Root**:\n   - Start the DFS traversal from the root and consider the root as a potential new tree.\n\n4. **Collect Results**:\n   - The `forest` list accumulates the roots of all the trees in the resulting forest.\n\n### Complexity\n- **Time Complexity**: \n  - The time complexity is O(n), where \\(n\\) is the number of nodes in the tree. This is because we visit each node exactly once in the DFS traversal.\n\n- **Space Complexity**: \n  - The space complexity is O(n) in the worst case, which accounts for the space used by the recursion stack in a skewed tree. Additionally, the space used to store the `to_delete` set and the `forest` list is O(n).\n\n```python\n# Definition for a binary tree node.\n# class TreeNode:\n#     def __init__(self, val=0, left=None, right=None):\n#         self.val = val\n#         self.left = left\n#         self.right = right\n\nclass Solution:\n    def delNodes(self, root: Optional[TreeNode], to_delete: List[int]) -> List[TreeNode]:\n        # Convert to_delete list to a set for O(1) lookups\n        to_delete_set = set(to_delete)\n        # List to store the roots of the resulting forest\n        forest = []\n        \n        def dfs(node, is_root):\n            if not node:\n                return None\n            root_deleted = node.val in to_delete_set\n            if is_root and not root_deleted:\n                forest.append(node)\n            # Recursively process the left and right children\n            node.left = dfs(node.left, root_deleted)\n            node.right = dfs(node.right, root_deleted)\n            return None if root_deleted else node\n        \n        # Start DFS traversal from the root\n        dfs(root, True)\n        return forest\n```\n
['Python3']