<p>Given a root node reference of a BST and a key, delete the node with the given key in the BST. Return <em>the <strong>root node reference</strong> (possibly updated) of the BST</em>.</p>

<p>Basically, the deletion can be divided into two stages:</p>

<ol>
	<li>Search for a node to remove.</li>
	<li>If the node is found, delete the node.</li>
</ol>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/04/del_node_1.jpg" style="width: 800px; height: 214px;" />
<pre>
<strong>Input:</strong> root = [5,3,6,2,4,null,7], key = 3
<strong>Output:</strong> [5,4,6,2,null,null,7]
<strong>Explanation:</strong> Given key to delete is 3. So we find the node with value 3 and delete it.
One valid answer is [5,4,6,2,null,null,7], shown in the above BST.
Please notice that another valid answer is [5,2,6,null,4,null,7] and it&#39;s also accepted.
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/04/del_node_supp.jpg" style="width: 350px; height: 255px;" />
</pre>

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

<pre>
<strong>Input:</strong> root = [5,3,6,2,4,null,7], key = 0
<strong>Output:</strong> [5,3,6,2,4,null,7]
<strong>Explanation:</strong> The tree does not contain a node with value = 0.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> root = [], key = 0
<strong>Output:</strong> []
</pre>

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

<ul>
	<li>The number of nodes in the tree is in the range <code>[0, 10<sup>4</sup>]</code>.</li>
	<li><code>-10<sup>5</sup> &lt;= Node.val &lt;= 10<sup>5</sup></code></li>
	<li>Each node has a <strong>unique</strong> value.</li>
	<li><code>root</code> is a valid binary search tree.</li>
	<li><code>-10<sup>5</sup> &lt;= key &lt;= 10<sup>5</sup></code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Could you solve it with time complexity <code>O(height of tree)</code>?</p>

/**
 * 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:
void dfs(TreeNode* root,vector<int>& nodes,int key){
    if(root==nullptr) return;
    if(root->val!=key) nodes.push_back(root->val);
    dfs(root->left,nodes,key);
    dfs(root->right,nodes,key);
}
TreeNode* create(vector<int>& nodes,int s,int e){
    int mid=(s+e)/2;
    TreeNode* ans=new TreeNode(nodes[mid]);
    if(s<mid){
        ans->left=create(nodes,s,mid-1);
    }
    else{
        ans->left=nullptr;
    }
    if(e>mid){
        ans->right=create(nodes,mid+1,e);
    }
    else {
        ans->right=nullptr;
    }
    return ans;
}
    TreeNode* deleteNode(TreeNode* root, int key) {
        vector<int>nodes;
        dfs(root,nodes,key);
        if(nodes.size()==0) return nullptr;
        sort(nodes.begin(),nodes.end());
        return create(nodes,0,nodes.size()-1);
    }
};
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