<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 {
private:
    void inorderTraversal(TreeNode* node,vector<int>&temp){
      if(node == NULL){
        return;
      }
      inorderTraversal(node->left,temp);
      temp.push_back(node->val);
      inorderTraversal(node->right,temp);
    }
    int binarySearch(vector<int>& numbers, int target, int start) {
      int low = start, high = numbers.size() - 1;
      while (low <= high) {
        int mid = low + (high - low) / 2;
        if (numbers[mid] == target) {
          return mid;
        } 
        else if (numbers[mid] > target) {
          high = mid - 1;
        }
        else {
          low = mid + 1;
        }
      }
      return -1;
    }
public:
    bool findTarget(TreeNode* root, int k) {
      vector<int>temp;
      inorderTraversal(root,temp);
      sort(temp.begin(),temp.end());
      for (int i = 0; i < temp.size(); i++) {
        int num = k - temp[i];
        int ans = binarySearch(temp, num, i + 1);  
        if (ans != -1) {
            return true;
        }
      }
      return false;
    }
};
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