1483. 树节点的第 K 个祖先
提示
困难
150
相关企业
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
Solution
第一次接触LCA算法。思想有点类似快速幂,先进行离线处理,之后按照位的思想进行运算即可。
class TreeAncestor:
def __init__(self, n: int, parent: List[int]):
m = n.bit_length() - 1
pa = [[p] + [-1] * m for p in parent]
for i in range(m):
for j in range(n):
if (p := pa[j][i]) != -1:
pa[j][i + 1] = pa[p][i]
self.pa = pa
def getKthAncestor(self, node: int, k: int) -> int:
bit = 0
while k and node != -1:
if k & 1:
node = self.pa[node][bit]
bit += 1
k >>= 1
return node
# Your TreeAncestor object will be instantiated and called as such:
# obj = TreeAncestor(n, parent)
# param_1 = obj.getKthAncestor(node,k)
标签:node,int,6.12,每日,节点,pa,TreeAncestor,getKthAncestor,leetcode
From: https://blog.51cto.com/u_15763108/6461435