1305. All Elements in Two Binary Search Trees
Medium
Input: root1 = [2,1,4], root2 = [1,0,3]
Output: [0,1,1,2,3,4]Input: root1 = [1,null,8], root2 = [8,1]
Output: [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]:
list1 = []
list2 = []
self.traverse(root1,list1)
self.traverse(root2, list2)
return self.merge(list1, list2)
def merge(self, list1, list2):
result = []
length1 = len(list1)
length2 = len(list2)
i = 0
j = 0
while i < length1 and j < length2:
if list1[i] <= list2[j]:
result.append(list1[i])
i += 1
else:
result.append(list2[j])
j += 1
if i < length1:
result.extend(list1[i:])
if j < length2:
result.extend(list2[j:])
return result
def traverse(self, root , arr):
if root is None:
return
self.traverse(root.left, arr)
if root is not None:
arr.append(root.val)
self.traverse(root.right, arr)Last updated

