<p>Given the <code>root</code> of a binary search tree, and an integer <code>k</code>, return <em>the</em> <code>k<sup>th</sup></code> <em>smallest value (<strong>1-indexed</strong>) of all the values of the nodes in the tree</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/28/kthtree1.jpg" style="width: 212px; height: 301px;" />
<pre>
<strong>Input:</strong> root = [3,1,4,null,2], k = 1
<strong>Output:</strong> 1
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/28/kthtree2.jpg" style="width: 382px; height: 302px;" />
<pre>
<strong>Input:</strong> root = [5,3,6,2,4,null,null,1], k = 3
<strong>Output:</strong> 3
</pre>

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

<ul>
	<li>The number of nodes in the tree is <code>n</code>.</li>
	<li><code>1 &lt;= k &lt;= n &lt;= 10<sup>4</sup></code></li>
	<li><code>0 &lt;= Node.val &lt;= 10<sup>4</sup></code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> If the BST is modified often (i.e., we can do insert and delete operations) and you need to find the kth smallest frequently, how would you optimize?</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 processNode(TreeNode* root, vector<int>&ans)
    {
        ans.push_back(root->val);
    }
    void inorder(TreeNode* root, vector<int>&ans)
    {
        if(root->left!=NULL)
        {
             inorder(root->left,ans);
        }
        processNode(root, ans);
    
        if(root->right!=NULL)
        {
             inorder(root->right,ans);
        }

    }
    int kthSmallest(TreeNode* root, int k) 
    {

        
        TreeNode* curr = root;
        TreeNode* prev = NULL;
        stack<TreeNode*>s;
        s.push(root);
        vector<int>ans;

        while(true) 
        {
            if(curr!=NULL)
            {
                s.push(curr);
                curr = curr->left;
            }
            else
            {
                if(s.empty()){break;}
             curr = s.top();
                ans.push_back(curr->val);
                s.pop();
                curr = curr->right;
                
            }

        }
        // vector<int>ans;
        // inorder(root, ans);

        return ans[k-1];
        
    }
};
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