# 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 ith node. The root of the tree is node 0. Find the kth ancestor of a given node.

The kth ancestor of a tree node is the kth 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 kth 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.

• 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(int n, vector<int>& parent) {
p = vector<vector<int>>(n, vector<int>(18, -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];
}
}
}

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

private:
vector<vector<int>> p;
};

/**
* 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:
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)


• type TreeAncestor struct {
p [][18]int
}

func Constructor(n int, parent []int) TreeAncestor {
p := make([][18]int, n)
for i, fa := range parent {
p[i][0] = fa
for j := 1; j < 18; j++ {
p[i][j] = -1
}
}
for j := 1; j < 18; j++ {
for i := range p {
if p[i][j-1] == -1 {
continue
}
p[i][j] = p[p[i][j-1]][j-1]
}
}
return TreeAncestor{p}
}

func (this *TreeAncestor) GetKthAncestor(node int, k int) int {
for i := 17; i >= 0; i-- {
if k>>i&1 == 1 {
node = this.p[node][i]
if node == -1 {
break
}
}
}
return node
}

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


• class TreeAncestor {
private p: number[][];

constructor(n: number, parent: number[]) {
const p = new Array(n).fill(0).map(() => new Array(18).fill(-1));
for (let i = 0; i < n; ++i) {
p[i][0] = parent[i];
}
for (let j = 1; j < 18; ++j) {
for (let i = 0; i < n; ++i) {
if (p[i][j - 1] === -1) {
continue;
}
p[i][j] = p[p[i][j - 1]][j - 1];
}
}
this.p = p;
}

getKthAncestor(node: number, k: number): number {
for (let i = 17; i >= 0; --i) {
if (((k >> i) & 1) === 1) {
node = this.p[node][i];
if (node === -1) {
break;
}
}
}
return node;
}
}

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


• public class TreeAncestor {
private int[][] p;

public TreeAncestor(int n, int[] parent) {
p = new int[n][];
for (int i = 0; i < n; i++) {
p[i] = new int[18];
for (int j = 0; j < 18; j++) {
p[i][j] = -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);
*/