144.binary-tree-preorder-traversal

Statement

• Difficulty: Easy
• Tag: `栈` `树` `深度优先搜索` `二叉树`

``````输入：root = [1,null,2,3]

``````

``````输入：root = []

``````

``````输入：root = [1]

``````

``````输入：root = [1,2]

``````

``````输入：root = [1,null,2]

``````

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

• Link: Binary Tree Preorder Traversal
• Difficulty: Easy
• Tag: `Stack` `Tree` `Depth-First Search` `Binary Tree`

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
``````