LeetCode in Python 1110. Delete Nodes And Return Forest

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.

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def delNodes(self, root, to_delete):
        """
        :type root: TreeNode
        :type to_delete: List[int]
        :rtype: List[TreeNode]
        """
        res = []
        self.helper(root, to_delete, res, True)
        return res
    
    def helper(self, root, to_delete, res, is_root):
        if not root:
            return None
        if root.val in to_delete:
            self.helper(root.left, to_delete, res, True)
            self.helper(root.right, to_delete, res, True)
            return None   
        else:
            if is_root:
                res.append(root)
                
            root.left = self.helper(root.left, to_delete, res, False)
            root.right = self.helper(root.right, to_delete, res, False)
            return root
  
        

第一遍没看题目return type，直接层次遍历了得到val了，注意到题目是返回List[TreeNode]之后重新写了，太久没刷二叉树题，连节点置空的操作都要忘了，第一次直接root=None还百思不得其解……正解是将父节点的left或right置空才是真的置空，此处需回忆一下指针的作用：指针指向内存的空间，是一个地址值，将指针置空并没有真的将指向关系删除，只有对待删除节点的父节点操作才是真正将这个节点从树结构中移除。

剩下的就比较常规了，和二叉树的dfs遍历类似。根据当前节点是否为根节点(is_root)来判断当前节点是否是一个解。