Spirit God - multiplied lcm template on the tree

class TreeAncestor:
    def __init__(self, edges: List[List[int]]):
        n = len(edges) + 1
        m = n.bit_length()
        g = [[] for _ in range(n)]
        for x, y in edges:  # 节点编号从 0 开始
            g[x].append(y)
            g[y].append(x)

        depth = [0] * n
        pa = [[-1] * m for _ in range(n)]
        def dfs(x: int, fa: int) -> None:
            pa[x][0] = fa
            for y in g[x]:
                if y != fa:
                    depth[y] = depth[x] + 1
                    dfs(y, x)
        dfs(0, -1)

        for i in range(m - 1):
            for x in range(n):
                if (p := pa[x][i]) != -1:
                    pa[x][i + 1] = pa[p][i]
        self.depth = depth
        self.pa = pa

    def get_kth_ancestor(self, node: int, k: int) -> int:
        for i in range(k.bit_length()):
            if (k >> i) & 1:  # k 二进制从低到高第 i 位是 1
                node = self.pa[node][i]
        return node

    # 返回 x 和 y 的最近公共祖先(节点编号从 0 开始)
    def get_lca(self, x: int, y: int) -> int:
        if self.depth[x] > self.depth[y]:
            x, y = y, x
        # 使 y 和 x 在同一深度
        y = self.get_kth_ancestor(y, self.depth[y] - self.depth[x])
        if y == x:
            return x
        for i in range(len(self.pa[x]) - 1, -1, -1):
            px, py = self.pa[x][i], self.pa[y][i]
            if px != py:
                x, y = px, py  # 同时上跳 2**i 步
        return self.pa[x][0]

作者:灵茶山艾府
链接:https://leetcode.cn/problems/kth-ancestor-of-a-tree-node/solutions/2305895/mo-ban-jiang-jie-shu-shang-bei-zeng-suan-v3rw/
来源:力扣(LeetCode)
著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。

Guess you like

Origin blog.csdn.net/weixin_40986490/article/details/131333090