Description
https://leetcode.com/problems/all-elements-in-two-binary-search-trees/
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 = [0,-10,10], root2 = [5,1,7,0,2] Output: [-10,0,0,1,2,5,7,10]
Example 3:
Input: root1 = [], root2 = [5,1,7,0,2] Output: [0,1,2,5,7]
Example 4:
Input: root1 = [0,-10,10], root2 = [] Output: [-10,0,10]
Example 5:
Input: root1 = [1,null,8], root2 = [8,1] Output: [1,1,8,8]
Constraints:
- Each tree has at most
5000nodes. - Each node’s value is between
[-10^5, 10^5].
Explanation
Use in order traverse on both trees to get two ascending order lists. Then merge two lists into one ascending order list.
Python Solution
# 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]:
results1 = []
self.inorder_traverse(root1, results1)
results2 = []
self.inorder_traverse(root2, results2)
results = []
i = 0
j = 0
while i < len(results1) and j < len(results2):
if results1[i] < results2[j]:
results.append(results1[i])
i += 1
else:
results.append(results2[j])
j += 1
while i < len(results1):
results.append(results1[i])
i += 1
while j < len(results2):
results.append(results2[j])
j += 1
return results
def inorder_traverse(self, root, results):
if not root:
return
self.inorder_traverse(root.left, results)
results.append(root.val)
self.inorder_traverse(root.right, results)
- Time Complexity: O(N).
- Space Complexity: O(N).