Leetcode #2689: Extract Kth Character From The Rope Tree
In this guide, we solve Leetcode #2689 Extract Kth Character From The Rope Tree in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given the root of a binary tree and an integer k. Besides the left and right children, every node of this tree has two other properties, a string node.val containing only lowercase English letters (possibly empty) and a non-negative integer node.len.
Quick Facts
- Difficulty: Easy
- Premium: Yes
- Tags: Tree, Depth-First Search, Binary Tree
Intuition
We need to explore a structure deeply before backing up, which suits DFS.
DFS keeps local context on the call stack and is easy to implement recursively.
Approach
Define a recursive DFS that carries the necessary state.
Combine child results as the recursion unwinds.
Steps:
- Define a recursive DFS with state.
- Visit children and combine results.
- Return the final aggregation.
Example
Input: root = [10,4,"abcpoe","g","rta"], k = 6
Output: "b"
Explanation: In the picture below, we put an integer on internal nodes that represents node.len, and a string on leaf nodes that represents node.val.
You can see that S[root] = concat(concat("g", "rta"), "abcpoe") = "grtaabcpoe". So S[root][5], which represents 6th character of it, is equal to "b".
Python Solution
# Definition for a rope tree node.
# class RopeTreeNode(object):
# def __init__(self, len=0, val="", left=None, right=None):
# self.len = len
# self.val = val
# self.left = left
# self.right = right
class Solution:
def getKthCharacter(self, root: Optional[object], k: int) -> str:
def dfs(root):
if root is None:
return ""
if root.len == 0:
return root.val
return dfs(root.left) + dfs(root.right)
return dfs(root)[k - 1]
Complexity
The time complexity is O(V+E). The space complexity is O(V).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.