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# 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:**

![](https://assets.leetcode.com/uploads/2019/07/01/screen-shot-2019-07-01-at-53836-pm.png)

```
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`.

```java
class Solution {
    List<TreeNode> ans;
    Set<Integer> set;

    public List<TreeNode> delNodes(TreeNode root, int[] to_delete) {
        ans = new ArrayList<>();
        set = new HashSet<>();
        for (int x : to_delete)
            set.add(x);
        helper(root);
        if (root != null && !set.contains(root.val))
            ans.add(root);
        return ans;
    }

    public TreeNode helper(TreeNode root) {
        if (root == null)
            return null;
        if (set.contains(root.val)) {
            if (root.left != null && !set.contains(root.left.val))
                ans.add(root.left);
            if (root.right != null && !set.contains(root.right.val))
                ans.add(root.right);
            helper(root.left);
            helper(root.right);
            return null;
        } else {
            root.left = helper(root.left);
            root.right = helper(root.right);
            return root;
        }
    }
}
```