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144.binary-tree-preorder-traversal

Statement

Metadata

给你二叉树的根节点 root ,返回它节点值的 前序 遍历。

 

示例 1:

输入:root = [1,null,2,3]
输出:[1,2,3]

示例 2:

输入:root = []
输出:[]

示例 3:

输入:root = [1]
输出:[1]

示例 4:

输入:root = [1,2]
输出:[1,2]

示例 5:

输入:root = [1,null,2]
输出:[1,2]

 

提示:

  • 树中节点数目在范围 [0, 100]
  • -100 <= Node.val <= 100

 

进阶:递归算法很简单,你可以通过迭代算法完成吗?

Metadata

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?

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
from typing import List, Optional


class Solution:
    def __init__(self):
        self.res = []

    def dfs(self, rt: TreeNode) -> None:
        if not rt:
            return

        self.res.append(rt.val)
        self.dfs(rt.left)
        self.dfs(rt.right)

    def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        self.dfs(root)
        return self.res

最后更新: January 15, 2023
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