21. 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]]

Example 2:

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

Solution: (Using Postorder)

Delete operation works only on NULL pointer

class Solution
{
public:
    vector<TreeNode *> res;

    void del(TreeNode *&root, unordered_set<int> &s)
    {
        if (root == NULL)
        {
            return;
        }

        del(root->left, s);
        del(root->right, s);

        if (s.find(root->val) != s.end())
        {
            if (root->left)
            {
                res.push_back(root->left);
            }

            if (root->right)
            {
                res.push_back(root->right);
            }

            s.erase(s.find(root->val));
            root = NULL;
            delete (root);
        }

        return;
    }

    vector<TreeNode *> delNodes(TreeNode *root, vector<int> &to_delete)
    {

        unordered_set<int> s;
        for (int i = 0; i < to_delete.size(); i++)
        {
            s.insert(to_delete[i]);
        }

        del(root, s);
        if (root)
        {
            res.push_back(root);
        }

        return res;
    }
};

Time Complexity: O(n)