<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> result;
        if (!root) return result;

        int maxFreq = 0, curFreq = 0, prevVal = INT_MIN;
        
        // First Pass: Find the max frequency
        TreeNode* cur = root;
        while (cur) {
            if (!cur->left) {
                // Process current node
                if (cur->val == prevVal) {
                    curFreq++;
                } else {
                    curFreq = 1;
                }
                prevVal = cur->val;
                maxFreq = max(maxFreq, curFreq);
                cur = cur->right;
            } else {
                // Find predecessor
                TreeNode* pred = cur->left;
                while (pred->right && pred->right != cur) {
                    pred = pred->right;
                }

                if (!pred->right) {
                    // Make cur the right child of its predecessor
                    pred->right = cur;
                    cur = cur->left;
                } else {
                    // Revert the changes made in the tree structure
                    pred->right = nullptr;

                    // Process current node
                    if (cur->val == prevVal) {
                        curFreq++;
                    } else {
                        curFreq = 1;
                    }
                    prevVal = cur->val;
                    maxFreq = max(maxFreq, curFreq);

                    cur = cur->right;
                }
            }
        }

        // Reset variables for the second pass
        curFreq = 0;
        prevVal = INT_MIN;
        cur = root;

        // Second Pass: Collect all values with max frequency
        while (cur) {
            if (!cur->left) {
                // Process current node
                if (cur->val == prevVal) {
                    curFreq++;
                } else {
                    curFreq = 1;
                }
                if (curFreq == maxFreq) {
                    result.push_back(cur->val);
                }
                prevVal = cur->val;
                cur = cur->right;
            } else {
                // Find predecessor
                TreeNode* pred = cur->left;
                while (pred->right && pred->right != cur) {
                    pred = pred->right;
                }

                if (!pred->right) {
                    // Make cur the right child of its predecessor
                    pred->right = cur;
                    cur = cur->left;
                } else {
                    // Revert the changes made in the tree structure
                    pred->right = nullptr;

                    // Process current node
                    if (cur->val == prevVal) {
                        curFreq++;
                    } else {
                        curFreq = 1;
                    }
                    if (curFreq == maxFreq) {
                        result.push_back(cur->val);
                    }
                    prevVal = cur->val;
                    cur = cur->right;
                }
            }
        }

        return result;
    }
};

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