Delete Nodes And Return Forest

https://leetcode.com/problems/delete-nodes-and-return-forest/

Delete Nodes And Return Forest

如果爸爸被删了，儿子自然成为forest里面某个root的候选人，如果儿子的val不在被删除的val里面，儿子就是forest的root。

这里node的reference先被添加到最后结果，之后再对node的左右孩子进行recursion delete nodes操作。

# 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]:
        self.res = []
        def dfs(node,is_root):
            if not node:
                return None
            delete_curr = node.val in to_delete
            if is_root and not delete_curr:
                self.res.append(node)
            node.left = dfs(node.left,delete_curr)
            node.right = dfs(node.right,delete_curr)
            if delete_curr:
                return None
            return node
        curr_is_root = root.val not in to_delete
        dfs(root,curr_is_root)
        return self.res