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.
Input: root = [1,2,3,4,5,6,7], to_delete = [3,5] Output: [[1,2,null,4],,]
Input: root = [1,2,4,null,3], to_delete =  Output: [[1,2,4]]
- The number of nodes in the given tree is at most
- Each node has a distinct value between
to_delete.length <= 1000
to_deletecontains distinct values between
Traverse the tree, if the node value is in the delete value list, change the node to be null, also check if the node has sub tree or not if there is a subtree, then mark it as a separate disjoint tree.
# 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: TreeNode, to_delete: List[int]) -> List[TreeNode]: results =  if not root: return results to_delete = set(to_delete) self.helper(root, to_delete, results, True) return results def helper(self, root, to_delete, results, is_root): if not root: return None if root.val in to_delete: if root.left: left = self.helper(root.left, to_delete, results, True) if root.right: right = self.helper(root.right, to_delete, results, True) root = None return root else: left = self.helper(root.left, to_delete, results, False) right = self.helper(root.right, to_delete, results, False) root.left = left root.right = right if is_root: results.append(root) return root
- Time Complexity: O(N).
- Space Complexity: O(N).
One Thought to “LeetCode 1110. Delete Nodes And Return Forest”
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