Problem Description

In this problem, we are given a tree structure with n nodes that are numbered from 0 to n - 1. A tree is a hierarchical structure where each node can have one parent and potentially multiple children, but no cycles exist. The tree is defined through a parent array named parent, where parent[i] represents the parent of the ith node. The node with number 0 is the root of the tree and it does not have a parent. Our task is to determine the kth ancestor of a given node, where the kth ancestor is the kth node in the path from the given node to the root. To represent this tree structure and perform ancestry queries, we need to implement a TreeAncestor class with two methods:

1. TreeAncestor(int n, int[] parent): Constructor method that initializes the tree with the number of nodes (n) and the parent array (parent).

2. getKthAncestor(int node, int k): Method that returns the kth ancestor of the provided node (node). If the ancestor does not exist (i.e., the kth level ancestor is beyond the root), it should return -1.

Flowchart Walkthrough

For leetcode 1483. Kth Ancestor of a Tree Node, let's apply the algorithm using the Flowchart. Here's a methodical analysis:

Is it a graph?

• Yes: The problem is defined on a tree, which is a special kind of graph.

Is it a tree?

• Yes: By definition, since each node has exactly one parent and ultimately connects to a root node, the structure is a tree.

Since we established it is a tree and looking at the nature of the query (finding Kth ancestor), the Depth-First Search (DFS) pattern fits well. This is primarily because DFS efficiently traverses trees, which allows for both the discovery of nodes and recording of their parent relationships, essential for answering "Kth ancestor" queries. DFS can handle the tree traversal needed to set up and answer these queries effectively.

Conclusion: Based on the flowchart traversal and the problem specifics, using DFS is appropriate for setting up data structures (like ancestor matrices or parent vectors) used to resolve Kth ancestor queries efficiently in a tree structure.

Intuition

Finding the kth ancestor naively by following the parent pointers one by one can be very inefficient, especially when k is large. To optimize the query time, we can precompute some data that enables us to "jump" multiple nodes at once.

This optimization relies on the concept of "binary lifting", which essentially means that we can represent the jump of k nodes as a sum of powers of 2, since any number can be represented as a combination of powers of 2. With this idea, we can precompute for each node what the 1st, 2nd, 4th (... 2^i-th ...) ancestor is. We store this information in a two-dimensional list p, where p[i][j] represents the 2^j-th ancestor of node i.

The constructor method prepares the p table, where the first column (j = 0) is just the direct parent of each node. For larger jumps (when j > 0), we can find the 2^j-th ancestor by looking at the 2^(j-1)-th ancestor of the node that is already the 2^(j-1)-th ancestor of the current node -- effectively making a "double jump".

When we want to find the kth ancestor of a node, we look at the binary representation of k. For each bit that is set in k, we jump to an ancestor that is 2^i steps up from the current node, where i is the position of that bit. This method drastically reduces the number of steps needed to find the kth ancestor.

This binary lifting technique helps in reducing a potentially large number of steps to a manageable few, bounded by the number of bits used to represent the number k.

Learn more about Tree, Depth-First Search, Breadth-First Search and Binary Search patterns.

Solution Approach

The solution uses a method called binary lifting to compute and query the kth ancestor of a node in a tree efficiently. Here is an in-depth explanation of how the algorithm works and the solution is implemented:

Data Structure Initialization

• A 2D list self.p of size n x 18 is created to store the powers of two ancestors for each node.
  • Why 18? Because for the constraints generally found in such problems, 2^17 is usually enough to cover the height of any tree (since 2^17 is greater than 10^5, which is a common maximum limit for n).
  • self.p[i][j] holds the 2^j-th ancestor of node i. -1 is used to indicate that such an ancestor does not exist.
• The __init__ method fills up the self.p matrix.
  • First, it copies the parent array to self.p[i][0] for all nodes because the 0-th power of 2 corresponds to immediate parents.
  • Then for each node i, it iterates over j from 1 to 17 (inclusive), using previously computed ancestors (from the j-1 step) to calculate the j-th power of 2 ancestor.
  • If at any point it encounters a -1, it implies that further ancestors do not exist, and it breaks from the inner loop.

Querying Ancestors

• The getKthAncestor method computes the kth ancestor for a given node.
  • It loops through powers of two from 2^17 down to 2^0, examining the bits of k.
  • For each i in 17 down to 0, it checks if the i-th bit of k is set (k >> i & 1). If it is, it jumps to the ancestor that is 2^i steps up by accessing self.p[node][i].
  • If during this process the node becomes -1, the method breaks out of the loop, as it indicates that we have tried to query above the root Ancestor (above the height of the tree).
  • The method returns the current node after the loop ends, which represents the k-th ancestor if one exists, or -1 if it doesn't.

This solution efficiently calculates the k-th ancestor for any given node by making at most log(k) jumps, drastically reducing the time complexity compared to a naive approach which would potentially require k jumps.

Example Walkthrough

Let's say we have a tree with 5 nodes where the parent array is [-1, 0, 0, 1, 2] and we want to find the Kth ancestor of node 4.

1. First, we initialize the TreeAncestor class with n = 5 and the parent array. This initiates the data structure and computes all 2^i-th ancestors for each node.

2. The initialization will produce a self.p table that might look something like this after computing the ancestors:

   | Node | 1st(2^0) | 2nd(2^1) | 4th(2^2) | ... |
   |------|----------|----------|----------|-----|
   | 0    | -1       | -1       | -1       | ... |
   | 1    | 0        | -1       | -1       | ... |
   | 2    | 0        | -1       | -1       | ... |
   | 3    | 1        | 0        | -1       | ... |
   | 4    | 2        | 0        | -1       | ... |

3. Now let's find the 2nd ancestor of node 4. We call getKthAncestor(4, 2).

4. In binary, k = 2 is 10. This means we want to find the ancestor that is 2^1 (the second bit from right to left in the binary representation of k) steps up from our current node.

5. We start from node 4 and check the highest power of two within 2, which is 2^1, and find the ancestor 2 steps up:

   • Access self.p[4][1], which gives us node 0 as the 2^1-th ancestor of node 4.

6. Since there is no higher power of two within 2, we have reached our result. The 2nd ancestor of node 4 is node 0 as per our self.p table.

This walkthrough demonstrates how the binary lifting method is used to compute the kth ancestor of a node by looking up precomputed ancestors using binary steps, significantly reducing the number of computations needed.

Solution Implementation

Time and Space Complexity

The given code defines a data structure for finding the k-th ancestor of a node in a tree with a pre-processing step that builds a sparse table for ancestor queries, and a query process that traverses this table to find the k-th ancestor.

Time Complexity

• Pre-processing (__init__ method): The pre-processing step iterates over each of the n nodes and fills the sparse table called self.p. This table has 18 levels because 2^17 is the highest power of 2 that is below the potential maximum of n which may be up to 50000 (as per usual constraints on LeetCode problems) allowing us to reach any ancestor k < n with at most 17 jumps. Therefore, the time complexity for the pre-processing part is O(n * log n) since we have a nested loop where the outer loop runs for n and the inner loop for up to log n (which is 18 in this specific implementation assuming a reasonable upper bound for n).

• Query (getKthAncestor method): For a single query to find the k-th ancestor, the method uses a loop that potentially iterates 18 times (the maximum number of jumps we need to make). For each iteration, it performs an O(1) check to see if the k-th ancestor exists for the current power of 2. Hence, the time complexity per query is O(log n).

Space Complexity

• Pre-processing (__init__ method): The space complexity is O(n * logn). A sparse table (self.p) of size n x 18 is built, and each entry at self.p[i][j] represents the 2^j-th ancestor of node i. Since 18 is a constant which relates to the potential log n levels, the space complexity can be considered O(n * log n).

In conclusion, the pre-processing step has a time complexity of O(n * log n) and also a space complexity of O(n * log n). The query step has a time complexity of O(log n) per query. These complexities are assuming that bitwise operations and list indexing are O(1).

Learn more about how to find time and space complexity quickly using problem constraints.