题目：原题链接（中等）
标签：并查集、图
| 解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
|---|---|---|---|
| Ans 1 (Python) | O ( N ) O(N) O(N) | O ( N ) O(N) O(N) | 156ms (11.67%) |
| Ans 2 (Python) | |||
| Ans 3 (Python) |
解法一：
class DSU1:
def __init__(self, n: int):
self._n = n
self._array = [i for i in range(n)]
self._size = [1] * n
def find(self, i: int):
if self._array[i] != i:
self._array[i] = self.find(self._array[i])
return self._array[i]
def union(self, i: int, j: int):
i, j = self.find(i), self.find(j)
if self._size[i] >= self._size[j]:
self._array[j] = i
self._size[i] += self._size[j]
else:
self._array[i] = j
self._size[j] += self._size[i]
def group_num(self):
groups = set()
for i in range(len(self._array)):
if self._array[i] not in groups:
j = self.find(i)
if j not in groups:
groups.add(self.find(i))
return len(groups)
def __repr__(self):
return str(len(self._array)) + ":" + str(self._array)
class Solution:
def validateBinaryTreeNodes(self, n: int, leftChild: List[int], rightChild: List[int]) -> bool:
dsu = DSU1(n)
for i in range(n):
if leftChild[i] != -1:
if dsu.find(i) == dsu.find(leftChild[i]):
return False
else:
dsu.union(i, leftChild[i])
if rightChild[i] != -1:
if dsu.find(i) == dsu.find(rightChild[i]):
return False
else:
dsu.union(i, rightChild[i])
return dsu.group_num() == 1
