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‎en/1-1000/144-binary-tree-preorder-traversal.md

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# 144. Binary Tree Preorder Traversal - Best Practices of LeetCode Solutions
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LeetCode link: [144. Binary Tree Preorder Traversal](https://leetcode.com/problems/binary-tree-preorder-traversal), difficulty: **Easy**.
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## LeetCode description of "144. Binary Tree Preorder Traversal"
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Given the `root` of a binary tree, return _the preorder traversal of its nodes' values_.
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### [Example 1]
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**Input**: `root = [1,null,2,3]`
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**Output**: `[1,2,3]`
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**Explanation**:
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![](../../images/examples/144_1.png)
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### [Example 2]
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**Input**: `root = [1,2,3,4,5,null,8,null,null,6,7,9]`
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**Output**: `[1,2,4,5,6,7,3,8,9]`
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**Explanation**:
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![](../../images/examples/144_2.png)
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### [Example 3]
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**Input**: `root = []`
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**Output**: `[]`
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### [Example 4]
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**Input**: `root = [1]`
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**Output**: `[1]`
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### [Constraints]
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- The number of nodes in the tree is in the range `[0, 100]`.
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- `-100 <= Node.val <= 100`
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## Intuition
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Will be added later.
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## Steps
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Will be added later.
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## Complexity
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Recursive solution and iterative solution are equal in complexity.
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* Time: `O(N)`.
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* Space: `O(N)`.
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## Follow-up
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Recursive solution is trivial, could you do it iteratively?
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## Follow-up intuition
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Will be added later.
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## Python
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### Solution 1: Recursion
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```python
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class Solution:
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def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
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self.values = []
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self.traverse(root)
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return self.values
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def traverse(self, node):
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if node is None:
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return
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self.values.append(node.val)
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self.traverse(node.left)
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self.traverse(node.right)
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```
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### Solution 2: Iteration
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```python
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class Solution:
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def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
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values = []
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stack = []
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if root:
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stack.append(root)
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while stack:
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node = stack.pop()
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values.append(node.val)
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if node.right:
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stack.append(node.right)
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if node.left:
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stack.append(node.left)
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return values
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```
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## JavaScript
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```javascript
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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## Java
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```java
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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## C++
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```cpp
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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## C#
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```c#
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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## Go
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```go
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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## Ruby
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```ruby
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# Welcome to create a PR to complete the code of this language, thanks!
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```
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## C, Kotlin, Swift, Rust or other languages
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```
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// Welcome to create a PR to complete the code of this language, thanks!
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```

‎images/examples/144_1.png

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‎images/examples/144_2.png

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‎zh/1-1000/144-binary-tree-preorder-traversal.md

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# 144. 二叉树的前序遍历 - 力扣题解最佳实践
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力扣链接：[144. 二叉树的前序遍历](https://leetcode.cn/problems/binary-tree-preorder-traversal) ，难度：**容易**。
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## 力扣“144. 二叉树的前序遍历”问题描述
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给你二叉树的根节点 `root` ，返回它节点值的 **前序** 遍历。
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### [示例 1]
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**输入**: `root = [1,null,2,3]`
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**输出**: `[1,2,3]`
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**解释**:
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![](../../images/examples/144_1.png)
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### [示例 2]
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**输入**: `root = [1,2,3,4,5,null,8,null,null,6,7,9]`
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**输出**: `[1,2,4,5,6,7,3,8,9]`
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**解释**:
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![](../../images/examples/144_2.png)
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### [示例 3]
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**输入**: `root = []`
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**输出**: `[]`
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### [示例 4]
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**输入**: `root = [1]`
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**输出**: `[1]`
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### [约束]
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- 树中节点数目在范围 `[0, 100]` 内
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- `-100 <= Node.val <= 100`
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## 思路
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以后会被添加。
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## 步骤
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以后会被添加。
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## 复杂度
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* 时间：`O(N)`。
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* 空间：`O(N)`。
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## 进阶
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递归算法很简单，你可以通过迭代算法完成吗？
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## 进阶思路
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以后会被添加。
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## Python
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### Solution 1: Recursion
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```python
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class Solution:
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def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
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self.values = []
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self.traverse(root)
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return self.values
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def traverse(self, node):
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if node is None:
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return
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self.values.append(node.val)
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self.traverse(node.left)
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self.traverse(node.right)
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```
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### Solution 2: Iteration
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```python
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class Solution:
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def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
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values = []
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stack = []
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if root:
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stack.append(root)
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while stack:
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node = stack.pop()
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values.append(node.val)
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if node.right:
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stack.append(node.right)
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if node.left:
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stack.append(node.left)
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return values
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```
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## JavaScript
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```javascript
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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## Java
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```java
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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## C++
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```cpp
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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## C#
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```c#
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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## Go
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```go
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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## Ruby
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```ruby
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# Welcome to create a PR to complete the code of this language, thanks!
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```
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## C, Kotlin, Swift, Rust or other languages
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```
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// Welcome to create a PR to complete the code of this language, thanks!
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```

‎zh/1-1000/209-minimum-size-subarray-sum.md

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**输出**: `2`
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**解释**: `子数组 _[4,3]_ 是该条件下的长度最小的子数组。`
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**解释**: `子数组 [4,3] 是该条件下的长度最小的子数组。`
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### [示例 2]
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**输入**: `target = 4, nums = [1,4,4]`

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