题目：原题链接（中等）
标签：树、二叉树、二叉搜索树、深度优先搜索
| 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) | 52ms (98.02%) |
| Ans 2 (Python) | |||
| Ans 3 (Python) |
解法一（深度优先搜索）：
class Solution:
def __init__(self):
self.ans_node, self.ans_num = None, 0
def largestBSTSubtree(self, root: TreeNode) -> int:
if not root:
return 0
self.dfs(root)
return self.ans_num
def dfs(self, node):
# 返回值：是否为BST、包含的节点数量、最小值、最大值
if node.left and node.right:
left_bool, left_num, left_min, left_max = self.dfs(node.left)
right_bool, right_num, right_min, right_max = self.dfs(node.right)
if left_bool and right_bool and left_max < node.val < right_min:
if self.ans_num < left_num + right_num + 1:
self.ans_node, self.ans_num = node, left_num + right_num + 1
return True, left_num + right_num + 1, left_min, right_max
else:
return False, None, None, None
elif node.left:
left_bool, left_num, left_min, left_max = self.dfs(node.left)
if left_bool and left_max < node.val:
if self.ans_num < left_num + 1:
self.ans_node, self.ans_num = node, left_num + 1
return True, left_num + 1, left_min, node.val
else:
return False, None, None, None
elif node.right:
right_bool, right_num, right_min, right_max = self.dfs(node.right)
if right_bool and node.val < right_min:
if self.ans_num < right_num + 1:
self.ans_node, self.ans_num = node, right_num + 1
return True, right_num + 1, node.val, right_max
else:
return False, None, None, None
else:
if self.ans_num < 1:
self.ans_node, self.ans_num = node, 1
return True, 1, node.val, node.val
