<p>Given the <code>root</code> of a binary tree, each node in the tree has a distinct value.</p>

<p>After deleting all nodes with a value in <code>to_delete</code>, we are left with a forest (a disjoint union of trees).</p>

<p>Return the roots of the trees in the remaining forest. You may return the result in any order.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/07/01/screen-shot-2019-07-01-at-53836-pm.png" style="width: 237px; height: 150px;" />
<pre>
<strong>Input:</strong> root = [1,2,3,4,5,6,7], to_delete = [3,5]
<strong>Output:</strong> [[1,2,null,4],[6],[7]]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> root = [1,2,4,null,3], to_delete = [3]
<strong>Output:</strong> [[1,2,4]]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the given tree is at most <code>1000</code>.</li>
	<li>Each node has a distinct value between <code>1</code> and <code>1000</code>.</li>
	<li><code>to_delete.length &lt;= 1000</code></li>
	<li><code>to_delete</code> contains distinct values between <code>1</code> and <code>1000</code>.</li>
</ul>

# 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: Optional[TreeNode], to_delete: List[int]) -> List[TreeNode]:
        forest = []
        s = set(to_delete)

        def dfs(root)->TreeNode:
            if root is None:
                return root
            
            root.left = dfs(root.left)
            root.right = dfs(root.right)

            if root.val in s:
                if root.left is not None:
                    forest.append(root.left)
                if root.right is not None:
                    forest.append(root.right)
                return None
            
            return root

        if root is None:
            return []

        dfs(root)
        if root.val not in s:
            forest.append(root)
        return forest
16900