Leetcode #1483: Kth Ancestor of a Tree Node
In this guide, we solve Leetcode #1483 Kth Ancestor of a Tree Node 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 a tree with n nodes numbered from 0 to n - 1 in the form of a parent array parent where parent[i] is the parent of ith node. The root of the tree is node 0.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Bit Manipulation, Tree, Depth-First Search, Breadth-First Search, Design, Binary Search, Dynamic Programming
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
Input
["TreeAncestor", "getKthAncestor", "getKthAncestor", "getKthAncestor"]
[[7, [-1, 0, 0, 1, 1, 2, 2]], [3, 1], [5, 2], [6, 3]]
Output
[null, 1, 0, -1]
Explanation
TreeAncestor treeAncestor = new TreeAncestor(7, [-1, 0, 0, 1, 1, 2, 2]);
treeAncestor.getKthAncestor(3, 1); // returns 1 which is the parent of 3
treeAncestor.getKthAncestor(5, 2); // returns 0 which is the grandparent of 5
treeAncestor.getKthAncestor(6, 3); // returns -1 because there is no such ancestor
Python Solution
class TreeAncestor:
def __init__(self, n: int, parent: List[int]):
self.p = [[-1] * 18 for _ in range(n)]
for i, fa in enumerate(parent):
self.p[i][0] = fa
for j in range(1, 18):
for i in range(n):
if self.p[i][j - 1] == -1:
continue
self.p[i][j] = self.p[self.p[i][j - 1]][j - 1]
def getKthAncestor(self, node: int, k: int) -> int:
for i in range(17, -1, -1):
if k >> i & 1:
node = self.p[node][i]
if node == -1:
break
return node
# Your TreeAncestor object will be instantiated and called as such:
# obj = TreeAncestor(n, parent)
# param_1 = obj.getKthAncestor(node,k)
Complexity
The time complexity is O(log n) or O(n log n). The space complexity is , where is the number of nodes in the tree.
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.