# Solving Leetcode Interviews in Seconds with AI: Binary Tree Preorder Traversal


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "144" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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] Explanation:   Example 2:  Input: root = [1,2,3,4,5,null,8,null,null,6,7,9] Output: [1,2,4,5,6,7,3,8,9] Explanation:   Example 3:  Input: root = [] Output: []  Example 4:  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? 

	# Explanation
	*   **Iterative Approach:** Use a stack to simulate the recursive calls. Start by pushing the root onto the stack. While the stack is not empty, pop a node, add its value to the result, and then push its right and left children onto the stack (right child first to ensure left is processed before right).
*   **Handling Null Nodes:** Carefully handle null nodes by not pushing them onto the stack. This avoids unnecessary processing and potential errors.
*   **Order of Processing:** By pushing the right child before the left, the left child is processed first after the current node, maintaining the preorder traversal order (Node, Left, Right).

*   **Runtime Complexity:** O(N), where N is the number of nodes in the tree.
*   **Storage Complexity:** O(N) in the worst-case scenario (skewed tree). In a balanced tree, the space complexity will be O(log N).

	
	# Code
	```python
	from typing import List, Optional

# 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]:
        if not root:
            return []

        result = []
        stack = [root]

        while stack:
            node = stack.pop()
            result.append(node.val)

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

        return result
	```