<p>Given the <code>root</code> of a binary search tree (BST) with duplicates, return <em>all the <a href="https://en.wikipedia.org/wiki/Mode_(statistics)" target="_blank">mode(s)</a> (i.e., the most frequently occurred element) in it</em>.</p>

<p>If the tree has more than one mode, return them in <strong>any order</strong>.</p>

<p>Assume a BST is defined as follows:</p>

<ul>
	<li>The left subtree of a node contains only nodes with keys <strong>less than or equal to</strong> the node&#39;s key.</li>
	<li>The right subtree of a node contains only nodes with keys <strong>greater than or equal to</strong> the node&#39;s key.</li>
	<li>Both the left and right subtrees must also be binary search trees.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/11/mode-tree.jpg" style="width: 142px; height: 222px;" />
<pre>
<strong>Input:</strong> root = [1,null,2,2]
<strong>Output:</strong> [2]
</pre>

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

<pre>
<strong>Input:</strong> root = [0]
<strong>Output:</strong> [0]
</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>5</sup> &lt;= Node.val &lt;= 10<sup>5</sup></code></li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> Could you do that without using any extra space? (Assume that the implicit stack space incurred due to recursion does not count).
class Solution {
public:   
    vector<int> findMode(TreeNode* root) {
        vector<int> ans;
        int maxStreak = 0;
        int currStreak = 0;
        int currNum = 0;
        
        TreeNode* curr = root;
        while (curr != nullptr) {
            if (curr->left != nullptr) {
                // Find the friend
                TreeNode* friendNode = curr->left;
                while (friendNode->right != nullptr) {
                    friendNode = friendNode->right;
                }
                
                friendNode->right = curr;
                
                // Delete the edge after using it
                TreeNode* left = curr->left;
                curr->left = nullptr;
                curr = left;
            } else {
                // Handle the current node
                int num = curr->val;
                if (num == currNum) {
                    currStreak++;
                } else {
                    currStreak = 1;
                    currNum = num;
                }

                if (currStreak > maxStreak) {
                    ans = {};
                    maxStreak = currStreak;
                }

                if (currStreak == maxStreak) {
                    ans.push_back(num);
                }
                
                curr = curr->right;
            }
        }
        
        return ans;
    }
};
15000