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:
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
and1000
. to_delete.length <= 1000
to_delete
contains distinct values between1
and1000
.
# Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution: def delNodes(self, root: TreeNode, to_delete: List[int]) -> List[TreeNode]: forest = [] to_delete = set(to_delete) def buttom_up(root): if root: left, right = dfs(root.left), dfs(root.right) root.left, root.right = left, right if root.val in to_delete: if left: forest.append(left) if right: forest.append(right) else: return root dfs(root) return ([] if root.val in to_delete else [root]) + forest ######################################################### to_delete_set = set(to_delete) res = [] def top_down(root, is_root): if not root: return None root_deleted = root.val in to_delete_set if is_root and not root_deleted: res.append(root) root.left = helper(root.left, root_deleted) root.right = helper(root.right, root_deleted) return None if root_deleted else root top_down(root, True) return res