<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(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
//     vector<TreeNode*> ans;
//     void delNodeUtil(TreeNode* root, TreeNode* prev, unordered_set<int>& check){
//         if(!root) return;
//         delNodeUtil(root->left, root, check);
//         delNodeUtil(root->right, root, check);
//         if(check.count(root->val)){
//             if(root->left) ans.push_back(root->left);
//             if(root->right) ans.push_back(root->right);
//             if(prev && prev->left && prev->left->val == root->val)
//                 prev->left = NULL;
//             else if(prev && prev->right && prev->right->val == root->val)
//                 prev->right = NULL;
//         }
//     }
    
//     vector<TreeNode*> delNodes(TreeNode* root, vector<int>& to_delete) {
//         if(!root) return {};
//         unordered_set<int> check(to_delete.begin(), to_delete.end());
//         ans.clear();
//         if(!check.count(root->val)) ans.push_back(root);
//         delNodeUtil(root, NULL, check);
//         return ans;
//     }
    void delNode(TreeNode* root, TreeNode* prev, unordered_set<int> del, vector<TreeNode*>& ans){
        if(!root) return;
        delNode(root->left, root, del, ans);
        delNode(root->right, root, del, ans);
        
        if(del.count(root->val)){
            TreeNode *right = root->right,
                     *left = root->left;
            if(left) ans.push_back(left);
            if(right) ans.push_back(right);
            // root->left = NULL, root->right = NULL;
            if(prev && prev->left && prev->left->val == root->val)
                    prev->left = NULL;
            if(prev && prev->right && prev->right->val == root->val)
                    prev->right = NULL;
        }
    }
    
    vector<TreeNode*> delNodes(TreeNode* root, vector<int>& to_delete) {
        if(!root) return {};
        vector<TreeNode*> ans;
        unordered_set<int> st(to_delete.begin(), to_delete.end());
        if(!st.count(root->val))
            ans.push_back(root);
        delNode(root, NULL, st, ans);
        return ans;
    }
};
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