<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 solve(TreeNode* root , vector<int>&ans){
    if (root == NULL)return ;

    solve(root->left , ans);
    ans.push_back(root->val);
    solve(root->right , ans);

  }
  bool findTarget(TreeNode* root, int k) {
    vector<int>ans;
    solve(root , ans);

    int n = ans.size();
    int l = 0 , r = n-1;
    while(l<r){
        if (ans[l] + ans[r] == k){
            return true;
        }
        if (ans[l]+ans[r]<k){
            l++;
        }
        else {
            r--;
        }
    }
    return false;

  }
};
36000