1305-all-elements-in-two-binary-search-trees

Try it on leetcode

Description

Given two binary search trees root1 and root2, return a list containing all the integers from both trees sorted in ascending order.

 

Example 1:

Input: root1 = [2,1,4], root2 = [1,0,3]
Output: [0,1,1,2,3,4]

Example 2:

Input: root1 = [1,null,8], root2 = [8,1]
Output: [1,1,8,8]

 

Constraints:

• The number of nodes in each tree is in the range [0, 5000].
• -105 <= Node.val <= 105

Solution(Python)

# 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 inorder(self, root, lst):
        if not root:
            return
        self.inorder(root.left, lst)
        lst.append(root.val)
        self.inorder(root.right, lst)

    def getAllElements(self, root1: TreeNode, root2: TreeNode) -> List[int]:
        return self.Linear(root1, root2)

    """
    Time Complexity: O((m+n)*log(m+n))
    Space Complexity: O(max(m,n))
    """

    def bruteforce(self, root1: TreeNode, root2: TreeNode) -> List[int]:
        res = []
        self.inorder(root1, res)
        self.inorder(root2, res)

        res.sort()
        return res

    """
    Time Complexity: O(m+n)
    Space Complexity: O(m+n)
    """

    def Linear(self, root1: TreeNode, root2: TreeNode) -> List[int]:
        lst1, lst2 = [], []
        self.inorder(root1, lst1)
        self.inorder(root2, lst2)

        res = []
        i, j = 0, 0

        while i < len(lst1) and j < len(lst2):
            if lst1[i] < lst2[j]:
                res.append(lst1[i])
                i += 1
            else:
                res.append(lst2[j])
                j += 1
        while i < len(lst1):
            res.append(lst1[i])
            i += 1
        while j < len(lst2):
            res.append(lst2[j])
            j += 1
        return res