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1483. 树节点的第 K 个祖先

English Version

题目描述

给你一棵树，树上有 n 个节点，按从 0 到 n-1 编号。树以父节点数组的形式给出，其中 parent[i] 是节点 i 的父节点。树的根节点是编号为 0 的节点。

树节点的第 k 个祖先节点是从该节点到根节点路径上的第 k 个节点。

实现 TreeAncestor 类：

TreeAncestor（int n， int[] parent） 对树和父数组中的节点数初始化对象。
getKthAncestor(int node, int k) 返回节点 node 的第 k 个祖先节点。如果不存在这样的祖先节点，返回 -1 。

示例 1：

输入：
["TreeAncestor","getKthAncestor","getKthAncestor","getKthAncestor"]
[[7,[-1,0,0,1,1,2,2]],[3,1],[5,2],[6,3]]

输出：
[null,1,0,-1]

解释：
TreeAncestor treeAncestor = new TreeAncestor(7, [-1, 0, 0, 1, 1, 2, 2]);

treeAncestor.getKthAncestor(3, 1);  // 返回 1 ，它是 3 的父节点
treeAncestor.getKthAncestor(5, 2);  // 返回 0 ，它是 5 的祖父节点
treeAncestor.getKthAncestor(6, 3);  // 返回 -1 因为不存在满足要求的祖先节点

提示：

1 <= k <= n <= 5 * 10⁴
parent[0] == -1 表示编号为 0 的节点是根节点。
对于所有的 0 < i < n ，0 <= parent[i] < n 总成立
0 <= node < n
至多查询 5 * 10⁴ 次

解法

方法一：动态规划 + 倍增

题目要我们寻找节点 $node$ 的第 $k$ 个祖先节点，如果暴力求解，需要从 $node$ 开始向上遍历 $k$ 次，时间复杂度为 $O(k)$，显然会超时。

我们可以使用动态规划，结合倍增的思想来处理。

我们定义 $p[i][j]$ 表示节点 $i$ 的第 $2^j$ 个祖先节点，即 $p[i][j]$ 表示节点 $i$ 向上走 $2^j$ 步的节点。那么我们可以得到状态转移方程：

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

即：要想找到节点 $i$ 的第 $2^j$ 个祖先节点，我们可以先找到节点 $i$ 的第 $2^{j-1}$ 个祖先节点，然后再找到该节点的第 $2^{j-1}$ 个祖先节点即可。所以，我们要找到每个节点的距离为 $2^j$ 的祖先节点，直到达到树的最大高度。

之后对于每次查询，我们可以把 $k$ 拆成二进制的表示形式，然后根据二进制中 $1$ 的位置，累计向上查询，最终得到节点 $node$ 的第 $k$ 个祖先节点。

时间复杂度方面，初始化为 $O(n \times \log n)$，查询为 $O(\log n)$。空间复杂度 $O(n \times \log n)$。其中 $n$ 为树的节点数。

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Python3

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)

Java

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);
 */

C++

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);
 */

Go

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);
 */

TypeScript

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)
 */

C#

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);
 */

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1483. 树节点的第 K 个祖先

题目描述

解法

方法一：动态规划 + 倍增

Python3

Java

C++

Go

TypeScript

C#

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1483. 树节点的第 K 个祖先

题目描述

解法

方法一：动态规划 + 倍增

Python3

Java

C++

Go

TypeScript

C#