|
| 1 | +''' |
| 2 | +https://leetcode.com/problems/delete-nodes-and-return-forest/ |
| 3 | +Given the root of a binary tree, each node in the tree has a distinct value. |
| 4 | + |
| 5 | +After deleting all nodes with a value in to_delete, we are left with a forest (a disjoint union of trees). |
| 6 | + |
| 7 | +Return the roots of the trees in the remaining forest. You may return the result in any order. |
| 8 | + |
| 9 | +''' |
| 10 | +# Performance |
| 11 | +''' |
| 12 | +Runtime: 52 ms, faster than 99.34% of Python3 online submissions for Delete Nodes And Return Forest. |
| 13 | +Memory Usage: 12.8 MB, less than 100.00% of Python3 online submissions for Delete Nodes And Return Forest. |
| 14 | +''' |
| 15 | +''' |
| 16 | +n: number of node |
| 17 | +m: number of to_delete |
| 18 | +Runtime: O(n) |
| 19 | +Space: O(log n) or O(m) |
| 20 | +''' |
| 21 | + |
| 22 | + |
| 23 | +class Solution: |
| 24 | + def delNodes(self, root, to_delete): |
| 25 | + # Change to set to for better check runtime |
| 26 | + set_delete = set(to_delete) |
| 27 | + forests = [] |
| 28 | + # Only add a node in the forest when |
| 29 | + # 1) Its parent is in set_delete |
| 30 | + # 2) It's not in set_delete |
| 31 | + # dfs return False when 1) Node is none and 2) node is in set_delete |
| 32 | + # When that happen, set the parent link to that node to None |
| 33 | + |
| 34 | + def dfs(node, is_parent_in_delete): |
| 35 | + nonlocal set_delete, forests |
| 36 | + if not node: |
| 37 | + return False |
| 38 | + # Still traverse subsequent node but return False |
| 39 | + if node.val in set_delete: |
| 40 | + if node.left: |
| 41 | + dfs(node.left, True) |
| 42 | + if node.right: |
| 43 | + dfs(node.right, True) |
| 44 | + return False |
| 45 | + # Add the current node to the forest |
| 46 | + if is_parent_in_delete: |
| 47 | + forests.append(node) |
| 48 | + # Set the child to none when the child is in set_delete |
| 49 | + if node.left and not dfs(node.left, False): |
| 50 | + node.left = None |
| 51 | + if node.right and not dfs(node.right, False): |
| 52 | + node.right = None |
| 53 | + return True |
| 54 | + dfs(root, True) |
| 55 | + return forests |
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