1483. Kth Ancestor of a Tree Node

Description

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 i^th node. The root of the tree is node 0. Find the k^th ancestor of a given node.

The k^th ancestor of a tree node is the k^th node in the path from that node to the root node.

Implement the TreeAncestor class:

TreeAncestor(int n, int[] parent) Initializes the object with the number of nodes in the tree and the parent array.
int getKthAncestor(int node, int k) return the k^th ancestor of the given node node. If there is no such ancestor, return -1.

Example 1:

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

Constraints:

1 <= k <= n <= 5 * 10⁴
parent.length == n
parent[0] == -1
0 <= parent[i] < n for all 0 < i < n
0 <= node < n
There will be at most 5 * 10⁴ queries.

Solutions

Solution 1: Dynamic Programming + Binary Lifting

The problem asks us to find the $k$-th ancestor node of a node $node$. If we solve it by brute force, we need to traverse upwards from $node$ for $k$ times, which has a time complexity of $O(k)$ and will obviously exceed the time limit.

We can use dynamic programming combined with the idea of binary lifting to handle this.

We define $p[i][j]$ as the $2^j$-th ancestor node of node $i$, i.e., the node reached by moving $2^j$ steps upwards from node $i$. Then we can get the state transition equation:

$$ p[i][j] = p[p[i][j-1]][j-1] $$

That is, to find the $2^j$-th ancestor node of node $i$, we can first find the $2^{j-1}$-th ancestor node of node $i$, and then find the $2^{j-1}$-th ancestor node of this node. Therefore, we need to find the ancestor node of each node at a distance of $2^j$, until we reach the maximum height of the tree.

For each query later, we can decompose $k$ into its binary representation, and then according to the positions of $1$ in the binary, we accumulate the queries upwards, and finally get the $k$-th ancestor node of node $node$.

In terms of time complexity, the initialization is $O(n \times \log n)$, and the query is $O(\log n)$. The space complexity is $O(n \times \log n)$, where $n$ is the number of nodes in the tree.

Similar problems:

2836. Maximize Value of Function in a Ball Passing Game

Python3JavaC++GoTypeScriptC#

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)

class TreeAncestor {
    private int[][] p;

    public TreeAncestor(int n, int[] parent) {
        p = new int[n][18];
        for (var e : p) {
            Arrays.fill(e, -1);
        }
        for (int i = 0; i < n; ++i) {
            p[i][0] = parent[i];
        }
        for (int j = 1; j < 18; ++j) {
            for (int i = 0; i < n; ++i) {
                if (p[i][j - 1] == -1) {
                    continue;
                }
                p[i][j] = p[p[i][j - 1]][j - 1];
            }
        }
    }

    public int getKthAncestor(int node, int k) {
        for (int i = 17; i >= 0; --i) {
            if (((k >> i) & 1) == 1) {
                node = p[node][i];
                if (node == -1) {
                    break;
                }
            }
        }
        return node;
    }
}

/**
 * Your TreeAncestor object will be instantiated and called as such:
 * TreeAncestor obj = new TreeAncestor(n, parent);
 * int param_1 = obj.getKthAncestor(node,k);
 */

class TreeAncestor {
public:
    TreeAncestor(