<p>Given the <code>root</code> of a binary search tree and an integer <code>k</code>, return <code>true</code> <em>if there exist two elements in the BST such that their sum is equal to</em> <code>k</code>, <em>or</em> <code>false</code> <em>otherwise</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/21/sum_tree_1.jpg" style="width: 400px; height: 229px;" />
<pre>
<strong>Input:</strong> root = [5,3,6,2,4,null,7], k = 9
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/21/sum_tree_2.jpg" style="width: 400px; height: 229px;" />
<pre>
<strong>Input:</strong> root = [5,3,6,2,4,null,7], k = 28
<strong>Output:</strong> false
</pre>

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

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 10<sup>4</sup>]</code>.</li>
	<li><code>-10<sup>4</sup> &lt;= Node.val &lt;= 10<sup>4</sup></code></li>
	<li><code>root</code> is guaranteed to be a <strong>valid</strong> binary search tree.</li>
	<li><code>-10<sup>5</sup> &lt;= k &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {
public:
    void nodetoarray (TreeNode* &node, vector<int> &v){
        if(node == nullptr) return;
        v.push_back(node->val);
        nodetoarray(node->left, v);
        nodetoarray(node->right, v);
    }

    bool twoSum(vector<int> nums, int target){
        unordered_map<int,int> nums_map; // <nums ele,index>

        for(int i=0;i<nums.size();i++){
            int k = target - nums[i];

            if(nums_map.find(k) != nums_map.end()){
                return true;
            }
            nums_map[nums[i]] = i;
        }
        return false;
    }

    bool findTarget(TreeNode* root, int k) {
        vector<int> v;
        nodetoarray(root, v);
        int n = v.size();
        sort(v.begin(), v.end());
        return twoSum(v,k);
    }
};
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