<p>Given the <code>root</code> of a binary tree, each node in the tree has a distinct value.</p>

<p>After deleting all nodes with a value in <code>to_delete</code>, we are left with a forest (a disjoint union of trees).</p>

<p>Return the roots of the trees in the remaining forest. You may return the result in any order.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/07/01/screen-shot-2019-07-01-at-53836-pm.png" style="width: 237px; height: 150px;" />
<pre>
<strong>Input:</strong> root = [1,2,3,4,5,6,7], to_delete = [3,5]
<strong>Output:</strong> [[1,2,null,4],[6],[7]]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> root = [1,2,4,null,3], to_delete = [3]
<strong>Output:</strong> [[1,2,4]]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the given tree is at most <code>1000</code>.</li>
	<li>Each node has a distinct value between <code>1</code> and <code>1000</code>.</li>
	<li><code>to_delete.length &lt;= 1000</code></li>
	<li><code>to_delete</code> contains distinct values between <code>1</code> and <code>1000</code>.</li>
</ul>

/**
 * 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), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<TreeNode*> delNodes(TreeNode* root, vector<int>& to_delete) {
        vector<TreeNode*> result;
        //doDeletion(result, root, to_delete);

        root = doDeletion(result, root, to_delete);
        if (root != nullptr) {
            result.push_back(root);
        }

        return result;
    }

    TreeNode* doDeletion(vector<TreeNode *> &result, TreeNode* root, vector<int> toDelete) {
        if (root == nullptr) {
            return root;
        }

        root->left = doDeletion(result, root->left, toDelete);
        root->right = doDeletion(result, root->right, toDelete);

        bool shouldBeDeleted = false;
        for (int i = 0; i < toDelete.size(); i++) {
            if (toDelete[i] == root->val) {
                shouldBeDeleted = true;
                break;
            }
        }

        if (shouldBeDeleted) {
            if (root->left != nullptr) {
                result.push_back(root->left);
            }

            if (root->right != nullptr) {
                result.push_back(root->right);
            }
            
            return nullptr;
        } else {
            return root;
        }
    }
};
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