LeetCode 1483. Kth Ancestor of a Tree Node Solution in Java, C++, Python & More | Explanation + Code

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 × log n), and the query is O(log n). The space complexity is O(n × log n), where n is the number of nodes in the tree.

Similar problems:

• 2836. Maximize Value of Function in a Ball Passing Game

PythonJavaC++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)(code-box)

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); */(code-box)

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); */(code-box)

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); */(code-box)

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) */(code-box)

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); */(code-box)