[数据结构与算法]LeetCode 1110. Delete Nodes And Return Forest - 二叉树（Binary Tree）系列题10

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]]

Example 2:

Input: root = [1,2,4,null,3], to_delete = [3]
Output: [[1,2,4]]

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.

题目给定一棵二叉树和一个节点数组to_delete，要求从树中把to_delete中的节点删掉，这样就得一棵棵互不相连的小树。要求返回每棵小树的根节点。

基本思路是利用前序遍历处理每个节点。对于一个节点无外乎两种情况，在或不在to_delete中。为了便于查找节点是否在to_delete中，可以把to_delete中的节点放入一个set中；另外为了便于处理节点之间的连接关系，在遍历过程中还需引入当前节点的父节点

1）节点在to_delete中，说明该节点需要被删掉，如果该节点的父节点不为空，则要把父节点的左子节点或者右子节点赋为空（取决于当前节点是其父节点的哪个子节点）。由于该节点被删掉了，所以在继续遍历其子树时，其子节点的父节点就应该为空。

2）节点不在to_delete中，说明该节点要保留，这时需要看该节点是否为一棵新小树的根节点，这取决于其父节点。如果父节点为空（即被删掉了），那么该节点为一棵新小树的根节点，否则就是一个普通的中间节点，继续往下遍历。

class Solution:
    def delNodes(self, root: Optional[TreeNode], to_delete: List[int]) -> List[TreeNode]:
        res = []
        delSet = set(to_delete)
        
        def helper(parent, node):
            if not node:
                return
            if node.val in delSet:
                if parent :
                    if parent.left == node:
                        parent.left = None
                    else:
                        parent.right = None
                parent = None
            else:
                if not parent:
                    res.append(node)
                parent = node
            
            helper(parent, node.left)
            helper(parent, node.right)
        helper(None, root)
        
        return res