<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]:
        to_delete = set(to_delete)
        res = []
        q = deque([root])
        while q:
            curr = q.popleft()
            if curr.left:
                q.append(curr.left)
                if curr.left.val in to_delete:
                    curr.left = None
            if curr.right:
                q.append(curr.right)
                if curr.right.val in to_delete:
                    curr.right = None
            if curr.val in to_delete:
                if curr.left:
                    res.append(curr.left)
                if curr.right:
                    res.append(curr.right)
        if root.val not in to_delete:
            res.append(root)
        return res

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