<p>Given two binary search trees <code>root1</code> and <code>root2</code>, return <em>a list containing all the integers from both trees sorted in <strong>ascending</strong> order</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/12/18/q2-e1.png" style="width: 457px; height: 207px;" />
<pre>
<strong>Input:</strong> root1 = [2,1,4], root2 = [1,0,3]
<strong>Output:</strong> [0,1,1,2,3,4]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/12/18/q2-e5-.png" style="width: 352px; height: 197px;" />
<pre>
<strong>Input:</strong> root1 = [1,null,8], root2 = [8,1]
<strong>Output:</strong> [1,1,8,8]
</pre>

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

<ul>
	<li>The number of nodes in each tree is in the range <code>[0, 5000]</code>.</li>
	<li><code>-10<sup>5</sup> &lt;= Node.val &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {
public:
    static vector<int> getAllElements(TreeNode* root1, TreeNode* root2) {
        vector<int> res;
        res.reserve(5000 * 2);
        function<void(TreeNode*)> dfs = [&](TreeNode* node) {
            if (node == nullptr) {
                return;
            }
            if (node->left != nullptr) {
                dfs(node->left);
            }
            res.push_back(node->val);
            if (node->right != nullptr) {
                dfs(node->right);
            }
        };
        dfs(root1);
        auto mid = res.begin() + res.size();
        dfs(root2);
        ranges::inplace_merge(res.begin(), mid, res.end());
        return res;
    }
};
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