DELETE NODES AND RETURN FOREST | LEETCODE # 1110 | PYTHON DFS SOLUTION

[@StellasAdi18]

Not the easiest problem… feels easy the way you explained. Thanks for explaining it.

[@rsKayiira]

Great solution this is not easy for me but well explained. Could you please add LC 825 Friends of Appropriate Ages

[@usamabaig5856]

How is T: O(n) + O(D), if for every node we are checking if its in to_delete then shouldn't it be O(n) * O(D), what am I missing?

[@Lord-vq5rh]

Nice Explanation

[@cpuneet98]

very helpful

[@MrZiyak99]

hey so if node.left.val is in to delete why do we set it as a root?

[@yousifshamoun5376]

If the node.val is in to_delete then is_root will never be checked so we could have set it to anything

[@floriandubost5484]

Yes, as Yousif said, this is kind of redundant. The code below does not have this and also runs:
class Solution:
def delNodes(self, root: Optional[TreeNode], to_delete: List[int]) -> List[TreeNode]:
out = []
to_delete = set(to_delete)
def dfs(node, is_root):
if not node:
return
if node.val in to_delete:
dfs(node.left, True)
dfs(node.right, True)
else:
dfs(node.left,False)
dfs(node.right,False)
if node.left and node.left.val in to_delete:
node.left = None
if node.right and node.right.val in to_delete:
node.right = None
if is_root:
out.append(node)
dfs(root, True)
return out

[@ryantomlinson9426]

+1 to this, it's a bit confusing setting that to root, but no matter what you set it to it still passes. I'd set it to False just for clarity.

[@somethingNewAndGood]

@@floriandubost5484 This is an awesome solution! A bit cleaner as well.

[@averychen4633]

@@floriandubost5484 Thank you. I was so confused by the code in the video. Your response is clearer!

[@seungminchoi8381]

Do you know how to do this without modifying the given tree?

[@lavanyam3224]

I guess the problem expects us to modify the tree coz we've to return the roots after deleting certain nodes.