<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]:

        ans = []

        def help(node):
            # base case 
            if not node:
                return False
            
            # check left child 
            if help(node.left): node.left = None
            
            # check right child 
            if help(node.right)==True: node.right = None

            # check the node itself 
            if node.val in to_delete:
                if node.left:
                    ans.append(node.left)
                if node.right:
                    ans.append(node.right)
                return True
            return False
        
        if not help(root):
            ans.append(root)
            
        
        return ans





























        if not root:
            return []

        to_delete_set = set(to_delete)
        forest = []
        root = self.processNode(root, to_delete_set, forest)

        if root:
            forest.append(root)
        
        return forest

    
    def processNode(self, node, to_delete_set, forest):

        if not node:
            return None
        
        #handle the left child 
        node.left = self.processNode(node.left,to_delete_set, forest)
        
        #handle the right child 
        node.right = self.processNode(node.right, to_delete_set, forest)

        #handle the curr node
        #if node itself needs to be deleted
        if node.val in to_delete_set:
            if node.left:
                forest.append(node.left)
            if node.right:
                forest.append(node.right)
            return None
        
        return node





        
        
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