Given therootof a binary tree, each node in the tree has a distinct value.

After deletingall nodes with a value into_delete, we are left with a forest (adisjoint 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 most1000.
• Each node has a distinct value between1and1000.
• to_delete.length <= 1000
• to_deletecontains distinct values between1and1000.

刪點成林。題意是給一棵二叉樹和一個to_delete陣列，請你從樹中刪除to_delete中出現的節點，返回一個森林。

這道題BFS和DFS都能做，我這裡提供一個BFS的做法。首先用一個hashset儲存所有需要刪除的節點。接著開始遍歷樹的每一個節點，遍歷的方式還是跟層序遍歷類似。從queue中彈出節點的時候，要做如下幾件事情

• 如果這個節點是需要被刪除的節點，需要從結果集裡面刪除這個節點
• 如果被刪除的節點有孩子節點，需要把孩子節點加入結果集，因為他們沒有父節點了，他們現在是自己這個子樹的根節點了
• 當前節點如果有左孩子和右孩子，則分別加入queue繼續下面的遍歷，同時遍歷的時候檢查他們是否需要被刪除，如果需要，在把他們放入queue之後需要斷開他們和父節點的連線

時間O(n)

空間O(n)

Java實現

 1 class Solution {
 2     public List<TreeNode> delNodes(TreeNode root, int[] to_delete) {
 3         List<TreeNode> res = new
 
 ArrayList<>();
 4         Queue<TreeNode> queue = new LinkedList<>();
 5         queue.offer(root);
 6         res.add(root);
 7         HashSet<Integer> set = new HashSet<>();
 8         for (int node : to_delete) {
 9             set.add(node);
10         }
11         
12         while (!queue.isEmpty()) {
13             TreeNode cur = queue.poll();
14             if (set.contains(cur.val)) {
15                 res.remove(cur);
16                 if (cur.left != null) {
17                     res.add(cur.left);
18                 }
19                 if (cur.right != null) {
20                     res.add(cur.right);
21                 }
22             }
23             if (cur.left != null) {
24                 queue.offer(cur.left);
25                 if (set.contains(cur.left.val)) {
26                     cur.left = null;
27                 }
28             }
29             if (cur.right != null) {
30                 queue.offer(cur.right);
31                 if (set.contains(cur.right.val)) {
32                     cur.right = null;
33                 }
34             }
35         }
36         return res;
37     }
38 }

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