[LeetCode] 1110. Delete Nodes And Return Forest

[grandyang]

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.

这道题给了一棵二叉树，说了每个结点值均不相同，现在让删除一些结点，由于删除某些位置的结点会使原来的二叉树断开，从而会形成多棵二叉树，形成一片森林，让返回森林中所有二叉树的根结点。对于二叉树的题，十有八九都是用递归来做的，这道题也不例外，先来想一下这道题的难点在哪里，去掉哪些点会形成新树，显而易见的是，去掉根结点的话，左右子树若存在的话一定会形成新树，同理，去掉子树的根结点，也可能会形成新树，只有去掉叶结点时才不会生成新树，所以当前结点是不是根结点就很重要了，这个需要当作一个参数传入。由于需要知道当前结点是否需要被删掉，每次都遍历 to_delete 数组显然不高效，那就将其放入一个 HashSet 中，从而到达常数级的搜索时间。这样递归函数就需要四个参数，当前结点，是否是根结点的布尔型变量，HashSet，还有结果数组 res。在递归函数中，首先判空，然后判断当前结点值是否在 HashSet，用一个布尔型变量 deleted 来记录。若当前是根结点，且不需要被删除，则将这个结点加入结果 res 中。然后将左子结点赋值为对左子结点调用递归函数的返回值，右子结点同样赋值为对右子结点调用递归的返回值，最后判断当前结点是否被删除了，是的话返回空指针，否则就返回当前指针，这样的话每棵树的根结点都在递归的过程中被存入结果 res 中了，参见代码如下：

class Solution {
public:
    vector<TreeNode*> delNodes(TreeNode* root, vector<int>& to_delete) {
        vector<TreeNode*> res;
        unordered_set<int> st(to_delete.begin(), to_delete.end());
        helper(root, true, st, res);
        return res;
    }
    TreeNode* helper(TreeNode* node, bool is_root, unordered_set<int>& st, vector<TreeNode*>& res) {
        if (!node) return nullptr;
        bool deleted = st.count(node->val);
        if (is_root && !deleted) res.push_back(node);
        node->left = helper(node->left, deleted, st, res);
        node->right = helper(node->right, deleted, st, res);
        return deleted ? nullptr : node;
    }
};

Github 同步地址:

#1110

参考资料：

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

https://leetcode.com/problems/delete-nodes-and-return-forest/discuss/328860/Simple-Java-Sol

https://leetcode.com/problems/delete-nodes-and-return-forest/discuss/328853/JavaC%2B%2BPython-Recursion-Solution

LeetCode All in One 题目讲解汇总(持续更新中...)

[lld2006]

is_root 有点烦， 在discussion上看到一个利用指针reference的code， 感觉很棒， 比较容易实现。

class Solution {
public:
    vector<TreeNode*> delNodes(TreeNode* root, vector<int>& to_delete) {
       vector<TreeNode*> result;
      set<int> delete_values(to_delete.begin(), to_delete.end());
      post_order(root, delete_values, result);
      if(root) result.push_back(root);
      return result;
    }
    void post_order(TreeNode*& node, set<int>& delete_values, vector<TreeNode*>& result){
      if(!node) return ;
      int value = node->val;
      post_order(node->left, delete_values, result);
      post_order(node->right, delete_values, result);
      if(delete_values.find(value) != delete_values.end()){
        if(node->left) result.push_back(node->left);
        if(node->right) result.push_back(node->right);
        //delete node;
        node=nullptr;
      }
    }
};