LeetCode 1110. Delete Nodes And Return Forest (二叉树构建非常好的题目!)

  1. Delete Nodes And Return Forest
    Medium

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.

解法1:
个人感觉这是一道非常好的关于二叉树的构建的题目。
这题我感觉挺难的。我是参考的labuladong的做法。

  1. 只有当没有parent并且不被删除,就是res中一个新的TreeNode *
  2. 如果一个节点被删除了,我们不能直接返回NULL,因为还要处理其子节点。其子节点的hasParent都是false。
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), righ
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