Skip to main content
Talentd

Extract Kth Character From The Rope Tree - Solution & Explanation

EasyPremiumFree on FleetCodeTreeDepth-First SearchBinary Tree6 min readAsked at: Google
Practice this problem

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. There are two types of nodes in this tree:

  • Leaf: These nodes have no children, node.len = 0, and node.val is some non-empty string.
  • Internal: These nodes have at least one child (also at most two children), node.len > 0, and node.val is an empty string.

The tree described above is called a Rope binary tree. Now we define S[node] recursively as follows:

  • If node is some leaf node, S[node] = node.val,
  • Otherwise if node is some internal node, S[node] = concat(S[node.left], S[node.right]) and S[node].length = node.len.

Return k-th character of the string S[root].

Note: If s and p are two strings, concat(s, p) is a string obtained by concatenating p to s. For example, concat("ab", "zz") = "abzz".

 

Example 1:

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".

Example 2:

Input: root = [12,6,6,"abc","efg","hij","klm"], k = 3
Output: "c"
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("abc", "efg"), concat("hij", "klm")) = "abcefghijklm". So S[root][2], which represents the 3rd character of it, is equal to "c".

Example 3:

Input: root = ["ropetree"], k = 8
Output: "e"
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] = "ropetree". So S[root][7], which represents 8th character of it, is equal to "e".

 

Constraints:

  • The number of nodes in the tree is in the range [1, 103]
  • node.val contains only lowercase English letters
  • 0 <= node.val.length <= 50
  • 0 <= node.len <= 104
  • for leaf nodes, node.len = 0 and node.val is non-empty
  • for internal nodes, node.len > 0 and node.val is empty
  • 1 <= k <= S[root].length

Approach Overview

Problem Overview: A rope tree stores a long string across multiple nodes. Leaf nodes contain actual string fragments, while internal nodes store the length of the left subtree. The task is to return the k-th character of the full concatenated string without building the entire string explicitly.

Approach 1: Build Full String with DFS (O(n) time, O(n) space)

The straightforward approach performs a depth‑first traversal and concatenates all leaf strings in order. Traverse the tree using DFS, append each leaf node's val to a result buffer, then return the character at index k. This approach treats the rope like a normal binary tree and ignores the optimization provided by the rope structure.

While simple, this method defeats the purpose of a rope. Constructing the full string costs O(n) time and space where n is the total number of characters. It works for small inputs or when you already need the entire string, but it is inefficient when only a single character is required.

Approach 2: Navigate Using Rope Length Metadata (O(h) time, O(1) space)

The rope structure is designed to support efficient indexing. Each internal node stores len, representing the total length of the left subtree. This value tells you exactly where the split occurs in the logical string.

Start at the root and compare k with len. If k < len, the target character lies in the left subtree, so move left. Otherwise, move to the right subtree and update the index to k - len. Continue this process until reaching a leaf node. At that point, simply return node.val[k].

This method works like binary search over the concatenated string. Instead of scanning characters, you skip entire substrings using the stored lengths. The traversal depth is bounded by the tree height h, so the runtime is O(h) with constant extra space. The algorithm naturally follows the structure of a tree and can be implemented using either iteration or depth‑first search.

Recommended for interviews: The length‑guided traversal is the approach interviewers expect. It demonstrates that you understand how rope data structures enable fast indexing without reconstructing the entire string. Mentioning the brute‑force DFS solution shows baseline understanding, but the optimized navigation using len proves you recognize how the rope structure is intended to be used.

Solution

Code

PythonJavaC++Go

Python

Java

C++

Go

Try this approach in the editor →

Detailed Complexity Analysis

ApproachTimeSpaceWhen to Use
DFS Build Full StringO(n)O(n)Simple implementation or when the full concatenated string is required
Length-Guided Rope TraversalO(h)O(1)Best approach for indexing a rope tree without constructing the entire string

Video Solution

leetcode 2689. Extract Kth Character From The Rope Tree - recursion and comments on problem • Code-Yao • 165 views views

Frequently Asked Questions

Is Extract Kth Character From The Rope Tree easy or hard?
The problem is rated Easy because the traversal logic is straightforward once you understand the rope structure. The main insight is using the stored subtree lengths to skip entire substrings instead of reconstructing the full string.
Extract Kth Character From The Rope Tree Python/Java solution
The solution navigates the rope tree using the stored left-subtree length. In Python, Java, C++, or Go, repeatedly compare k with node.len, move left if k is smaller, otherwise move right and subtract node.len from k. When a leaf node is reached, return node.val[k].
How to solve Extract Kth Character From The Rope Tree in O(h)?
Start at the root and compare k with the node's left-subtree length value. If k is smaller, move to the left child. Otherwise move right and subtract the left length from k. Continue until reaching a leaf node, then return the character from the leaf's string at index k.
What is the best approach for Extract Kth Character From The Rope Tree?
The optimal approach uses the rope node's stored length metadata. Each internal node keeps the size of its left subtree. Compare k with this value to decide whether to move left or right, subtracting the left length when moving right. This finds the character in O(h) time where h is the height of the tree.
Is Extract Kth Character From The Rope Tree asked at Google/Amazon/Meta?
Rope structures and substring indexing problems appear in interviews at companies that work heavily with text processing or editors. Variations involving trees, substring storage, and indexed traversal are common at companies like Google and Meta when testing tree and data structure knowledge.
What data structure is used in Extract Kth Character From The Rope Tree?
The problem uses a rope, which is a specialized binary tree designed for efficient string manipulation. Leaf nodes store actual string segments, while internal nodes store the length of the left subtree to support fast indexing and concatenation operations.
What is the time complexity of Extract Kth Character From The Rope Tree?
The optimal traversal runs in O(h) time where h is the rope tree height, because each step moves down one level. Space complexity is O(1) if implemented iteratively. A brute-force DFS that builds the entire string requires O(n) time and O(n) space.

Ready to solve this problem?

Practice Extract Kth Character From The Rope Tree with our built-in code editor and test cases.

Practice on FleetCode
Previous

Binary Tree Inorder Traversal

Next

Unique Binary Search Trees II