1110. Delete Nodes And Return Forest

Tree dfs

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.

分析

给定一个二叉树的根节点 root,以及一个需要删除的节点值的列表 to_delete。你需要删除 to_delete 列表中所有值的节点。删除这些节点后,可能会形成多个独立的二叉树(一个森林)。你需要返回这个森林中所有树的根节点列表。

from typing import List, Optional

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

class Solution:
    def deleteNodes(self, root: Optional[TreeNode], to_delete: List[int]) -> List[TreeNode]:
        # 用于存储森林中所有树的根节点
        forest = []
        # 将需要删除的值放入一个集合中,方便快速查找
        to_delete_set = set(to_delete)

        def dfs(node: Optional[TreeNode], is_root: bool) -> Optional[TreeNode]:
            """
            深度优先搜索遍历二叉树,删除指定节点并识别新的根节点。

            Args:
                node: 当前遍历的节点。
                is_root: 一个布尔值,表示当前节点是否是某个树的根节点。

            Returns:
                如果当前节点被删除,则返回 None;否则返回当前节点。
            """
            if not node:
                return None

            # 判断当前节点是否需要删除
            deleted = node.val in to_delete_set
            # 递归处理左右子树,并更新当前节点的左右子节点。
            # 如果子节点所在的子树的根被删除,那么当前节点指向该子节点的指针会变成 None。
            node.left = dfs(node.left, deleted)
            node.right = dfs(node.right, deleted)

            # 如果当前节点是某个树的根节点(即它的父节点被删除,或者它是原始的根节点且没有被删除),
            # 并且当前节点没有被删除,那么它就是森林中的一棵树的根,将其添加到 forest 列表中。
            if is_root and not deleted:
                forest.append(node)

            # 如果当前节点被删除,返回 None,这样它的父节点就不会再指向它了。
            # 如果当前节点没有被删除,返回当前节点。
            return None if deleted else node

        # 从根节点开始进行深度优先搜索,初始时根节点被认为是潜在的根节点 (is_root=True)
        dfs(root, True)
        return forest

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