144.Binary Tree Preorder Traversal
===

###### tags: `Easy`,`Tree`,`DFS`,`Stack`,`Binary Tree`

[144. Binary Tree Preorder Traversal](https://leetcode.com/problems/binary-tree-preorder-traversal/)

### 題目描述

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

### 範例

**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?

### 解答

#### Javascript
recursive

function preorderTraversal(root) {
  const result = [];
  if (root === null) return result;
  result.push(root.val);
  result.push(...preorderTraversal(root.left));
  result.push(...preorderTraversal(root.right));
  return result;
}

iterative

function preorderTraversal2(root) {
  const result = [];
  if (root === null) return result;
  const stack = [root];
  while (stack.length > 0) {
    const node = stack.pop();
    result.push(node.val);
    if (node.right !== null) stack.push(node.right);
    if (node.left !== null) stack.push(node.left);
  }

  return result;
}

> [name=Marsgoat] [time= Jan 9, 2023]

#### Python

class Solution:
    def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        if not root: return root
        
        traversal, stk = [], [root]

        while stk:
            node = stk.pop()
            traversal.append(node.val)

            if node.right:
                stk.append(node.right)
            if node.left:
                stk.append(node.left)

        return traversal

> [name=Ron Chen] [time= Jan 9, 2023]

### Reference

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