LeetCode 솔루션 분류
[9/8] 94. Binary Tree Inorder Traversal
본문
Easy
9501449Add to ListShareGiven the root of a binary tree, return the inorder traversal of its nodes' values.
Example 1:
Input: root = [1,null,2,3] Output: [1,3,2]
Example 2:
Input: root = [] Output: []
Example 3:
Input: root = [1] Output: [1]
Constraints:
- The number of nodes in the tree is in the range
[0, 100]. -100 <= Node.val <= 100
Follow up: Recursive solution is trivial, could you do it iteratively?
Accepted
1,714,193
Submissions
2,360,753
관련자료
-
링크
댓글 1
학부유학생님의 댓글
- 익명
- 작성일
# 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 inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
return self.inorderTraversal(root.left) + [root.val] + self.inorderTraversal(root.right) if root else []
class Solution:
def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
res = []
def inorder(node):
if not node: return
inorder(node.left)
res.append(node.val)
inorder(node.right)
inorder(root)
return res
