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# 1110.Delete-Nodes-And-Return-Forest

## 题目描述

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 and 1000.
to_delete.length <= 1000
to_delete contains distinct values between 1 and 1000.

## 代码

### Approach #1 Recursion

Time `O(N)` Space `O(H + N)`, where `H` is the height of tree.
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
Set<Integer> to_delete_set;
List<TreeNode> res;
public List<TreeNode> delNodes(TreeNode root, int[] to_delete) {
to_delete_set = new HashSet<>();
res = new ArrayList<>();
for (int i: to_delete) {