엔지니어 게시판
LeetCode 솔루션 분류

[1/9] 144. Binary Tree Preorder Traversal

컨텐츠 정보

본문

Given the root of a binary tree, return the preorder traversal of its nodes' values.

 

Example 1:

Input: root = [1,null,2,3]
Output: [1,2,3]

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
1.2M
<div class="text-label-2 dark:text-dark-label-2 text-xs" style="border: 0px solid; box-sizing: border-box; --tw-border-spacing-x:0; --tw-border-spacing-y:0; --tw-translate-x:0; --tw-translate-y:0; --tw-rotate:0; --tw-skew-x:0; --tw-skew-y:0; --tw-scale-x:1; --tw-scale-y:1; --tw-pan-x: ; --tw-pan-y: ; --tw-pinch-zoom: ; --tw-scroll-snap-strictness:proximity; --tw-ordinal: ; --tw-slashed-zero: ; --tw-numeric-figure: ; --tw-numeric-spacing: ; --tw-numeric-fraction: ; --tw-ring-inset: ; --tw-ring-offset-width:0px; --tw-ring-offset-color:#fff; --tw-ring-color:rgba(59,130,246,0.5); --tw-ring-offset-shadow:0 0 #0000; --tw-ring-shadow:0 0 #0
태그 , ,

관련자료

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 preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        res = []

        def dfs(node):
            if not node: return
            res.append(node.val)
            dfs(node.left)
            dfs(node.right)
        
        dfs(root)

        return res
전체 396 / 1 페이지
번호
제목
이름

최근글


인기글


새댓글


Stats


  • 현재 접속자 179 명
  • 오늘 방문자 4,327 명
  • 어제 방문자 5,332 명
  • 최대 방문자 11,134 명
  • 전체 회원수 1,048 명
알림 0