<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>

# 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: TreeNode, root2: TreeNode) -> List[int]:
        nums = []
        def traverse(root):
            if not root:
                return
            stack = [root]
            while stack:
                curr = stack.pop()
                if curr.left:
                    stack.append(curr.left)
                if curr.right:
                    stack.append(curr.right)
                nums.append(curr.val)
        traverse(root1)
        traverse(root2)
        return sorted(nums)


        
19800