Introduction

In this blog post, we will explore how to solve the LeetCode problem "257" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, 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 all root-to-leaf paths in any order. A leaf is a node with no children. Example 1: Input: root = [1,2,3,null,5] Output: ["1->2->5","1->3"] Example 2: Input: root = [1] Output: ["1"] Constraints: The number of nodes in the tree is in the range [1, 100]. -100 <= Node.val <= 100

Explanation

Here's the solution to the problem:

• Depth-First Search (DFS): Traverse the tree using DFS, keeping track of the current path from the root to the current node.
• Path Construction: When a leaf node is encountered, add the current path to the result list.

• Backtracking: As the DFS returns from each node, remove the node from the current path to explore other branches.

• Time Complexity: O(N), where N is the number of nodes in the tree. We visit each node once.

• Space Complexity: O(H), where H is the height of the tree, due to the recursion stack. In the worst case (skewed tree), H = N, leading to O(N) space. The space for storing the result paths is also O(N) in the worst-case scenario where every path is stored.

Code

    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 binaryTreePaths(self, root: Optional[TreeNode]) -> List[str]:
        paths = []

        def dfs(node: Optional[TreeNode], current_path: str):
            if node is None:
                return

            current_path += str(node.val)

            if node.left is None and node.right is None:
                paths.append(current_path)
                return

            current_path += "->"

            dfs(node.left, current_path)
            dfs(node.right, current_path)

        dfs(root, "")
        return paths