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getAllElements.py
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75 lines (56 loc) · 1.85 KB
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# 1305. 两个二叉搜索树中的所有元素
from typing import List
# 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]:
#
# def inorder(root):
# if not root:
# return []
# return inorder(root.left) + [root.val] + inorder(root.right)
#
# def merge(nums1, nums2):
# res = []
# m, n = len(nums1), len(nums2)
# i, j = 0, 0
# while i < m and j < n:
# if nums1[i] < nums2[j]:
# res.append(nums1[i])
# i += 1
# else:
# res.append(nums2[j])
# j += 1
# if i < m:
# res.extend(nums1[i:])
# else:
# res.extend(nums2[j:])
# return res
#
# return merge(inorder(root1), inorder(root2))
def getAllElements(self, root1: TreeNode, root2: TreeNode) -> List[int]:
def inorder(root):
if not root:
return []
return inorder(root.left) + [root.val] + inorder(root.right)
def merge(nums1, nums2):
res = []
m, n = len(nums1), len(nums2)
i, j = 0, 0
while i < m and j < n:
if nums1[i] < nums2[j]:
res.append(nums1[i])
i += 1
else:
res.append(nums2[j])
j += 1
if i < m:
res.extend(nums1[i:])
else:
res.extend(nums2[j:])
return res
return merge(inorder(root1), inorder(root2))