LeetCode 1305. 两棵二叉搜索树中的所有元素

示例 1：
输入：root1 = [2,1,4], root2 = [1,0,3]
输出：[0,1,1,2,3,4]
示例 2：
输入：root1 = [0,-10,10], root2 = [5,1,7,0,2]
输出：[-10,0,0,1,2,5,7,10]
示例 3：
输入：root1 = [], root2 = [5,1,7,0,2]
输出：[0,1,2,5,7]
示例 4：
输入：root1 = [0,-10,10], root2 = []
输出：[-10,0,10]

示例 5：
输入：root1 = [1,null,8], root2 = [8,1]
输出：[1,1,8,8]

# 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]:
        resList = []
        ## 普通二叉树遍历再排序
        def dfs(root:TreeNode):
            if root == None:
                return
            resList.append(root.val)
            if root.left:
                dfs(root.left)
            if root.right:
                dfs(root.right)
        dfs(root1)
        dfs(root2)
        resList.sort()
        return resList
if __name__ == '__main__':
    root1 = TreeNode(1)
    root1.left = TreeNode(2)
    root1.right = TreeNode(3)
    root2 = TreeNode(3)
    root2.left = TreeNode(4)
    root2.right = TreeNode(5)
    result = Solution().getAllElements(root1, root2)
    print(result)