How do you solve Kth Ancestor of a Tree Node on LeetCode?

Kth Ancestor of a Tree Node (LeetCode #1483) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) for preprocessing, O(log n) for each query time complexity and O(n log n) space complexity. Key insight: Using a sparse table allows for efficient ancestor retrieval by leveraging binary lifting.

What is the time complexity of Kth Ancestor of a Tree Node?

The optimal solution for Kth Ancestor of a Tree Node runs in O(n log n) for preprocessing, O(log n) for each query time and uses O(n log n) space. The preprocessing step takes O(n log n) due to filling the ancestor table, while each query takes O(log n) because we can jump through the tree in logarithmic steps.

What is the brute force solution for Kth Ancestor of a Tree Node?

The brute force approach for Kth Ancestor of a Tree Node is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves repeatedly moving up the tree from the given node to find the k-th ancestor. This is straightforward but inefficient for large k or many queries. The optimal Optimal Solution approach improves this to O(n log n) for preprocessing, O(log n) for each query.

What are the interview tips for Kth Ancestor of a Tree Node?

Always discuss your thought process and potential optimizations with the interviewer. If you start with a brute force solution, clearly explain why it is inefficient and how you can improve it. Practice explaining your code and the logic behind it as if teaching someone else.

What are common mistakes when solving Kth Ancestor of a Tree Node?

Not considering edge cases where the ancestor does not exist (return -1). Failing to optimize the solution when the number of queries is large.

#1483

Kth Ancestor of a Tree Node

Hard

Binary SearchDynamic ProgrammingBit ManipulationTreeDepth-First SearchBreadth-First SearchDesignBinary LiftingDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n) for preprocessing, O(log n) for each query
Space	O(1)	O(n log n)

💡

Intuition

Time O(n log n) for preprocessing, O(log n) for each querySpace O(n log n)

Using a sparse table approach allows us to precompute ancestors at different powers of two, enabling us to jump up the tree in logarithmic steps. This drastically reduces the time complexity for each query.

⚙️

Algorithm

3 steps

1Step 1: Precompute a table where table[i][j] represents the 2^j-th ancestor of node i.
2Step 2: For each query, use the precomputed table to find the k-th ancestor by breaking k down into powers of two.
3Step 3: If at any point the ancestor does not exist, return -1.

solution.py20 lines

1class TreeAncestor:
2    def __init__(self, n: int, parent: List[int]):
3        self.log = 1
4        while (1 << self.log) <= n:
5            self.log += 1
6        self.ancestor = [[-1] * self.log for _ in range(n)]
7        for i in range(n):
8            self.ancestor[i][0] = parent[i]
9        for j in range(1, self.log):
10            for i in range(n):
11                if self.ancestor[i][j - 1] != -1:
12                    self.ancestor[i][j] = self.ancestor[self.ancestor[i][j - 1]][j - 1]
13                
14    def getKthAncestor(self, node: int, k: int) -> int:
15        for j in range(self.log):
16            if k & (1 << j):
17                node = self.ancestor[node][j]
18                if node == -1:
19                    return -1
20        return node

ℹ

Complexity note: The preprocessing step takes O(n log n) due to filling the ancestor table, while each query takes O(log n) because we can jump through the tree in logarithmic steps.

1Using a sparse table allows for efficient ancestor retrieval by leveraging binary lifting.
2Understanding how to break down k into powers of two is crucial for optimizing the traversal.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.