All Elements in Two Binary Search Trees
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1300-1399/1305.All%20Elements%20in%20Two%20Binary%20Search%20Trees/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1300-1399/1305.All%20Elements%20in%20Two%20Binary%20Search%20Trees/images/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>
Tree; Depth-First Search; Binary Search Tree; Binary Tree; Sorting
/**
 * 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:
    vector<int> getAllElements(TreeNode* root1, TreeNode* root2) {
        vector<int> a, b, ans;
        dfs(root1, a);
        dfs(root2, b);

        int i = 0, j = 0;
        while (i < a.size() && j < b.size()) {
            if (a[i] <= b[j]) {
                ans.push_back(a[i++]);
            } else {
                ans.push_back(b[j++]);
            }
        }
        while (i < a.size()) {
            ans.push_back(a[i++]);
        }
        while (j < b.size()) {
            ans.push_back(b[j++]);
        }
        return ans;
    }

private:
    void dfs(TreeNode* root, vector<int>& nums) {
        if (root == nullptr) {
            return;
        }
        dfs(root->left, nums);
        nums.push_back(root->val);
        dfs(root->right, nums);
    }
};
