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.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def getAllElements(
        self, root1: Optional[TreeNode], root2: Optional[TreeNode]
    ) -> List[int]:
        def dfs(root: Optional[TreeNode], nums: List[int]) -> int:
            if root is None:
                return
            dfs(root.left, nums)
            nums.append(root.val)
            dfs(root.right, nums)

        a, b = [], []
        dfs(root1, a)
        dfs(root2, b)
        m, n = len(a), len(b)
        i = j = 0
        ans = []
        while i < m and j < n:
            if a[i] <= b[j]:
                ans.append(a[i])
                i += 1
            else:
                ans.append(b[j])
                j += 1
        while i < m:
            ans.append(a[i])
            i += 1
        while j < n:
            ans.append(b[j])
            j += 1
        return ans
