题目：原题链接（中等）
标签：设计、树、二叉树、二叉搜索树、栈
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) | 每个操作 : O ( l o g N ) | O ( l o g N ) | 516ms (87.50%) |
Ans 2 (Python) | |||
Ans 3 (Python) |
解法一（栈维护父节点栈）：
class BSTIterator:
def __init__(self, root: TreeNode):
self.root = root
self.start = False
self.parent = []
self.now = None
def hasNext(self) -> bool:
if not self.start:
return True
else:
if self.now.right:
return True
else:
val = self.now.val
for parent in self.parent:
if parent.val > val:
return True
if parent.right and parent.right.val > val:
return True
return False
def next(self) -> int:
if self.start:
if self.now.right:
self.parent.append(self.now)
self.now = self.now.right
while self.now.left:
self.parent.append(self.now)
self.now = self.now.left
return self.now.val
else:
val = self.now.val
while self.parent:
self.now = self.parent.pop()
if self.now.val > val:
return self.now.val
if self.now.right and self.now.right.val > val:
self.parent.append(self.now)
self.now = self.now.right
while self.now.left:
self.parent.append(self.now)
self.now = self.now.left
return self.now.val
return None
else:
self.start = True
self.now = self.root
while self.now.left:
self.parent.append(self.now)
self.now = self.now.left
return self.now.val
def hasPrev(self) -> bool:
if not self.start:
return False
elif self.now.left:
return True
else:
val = self.now.val
for parent in self.parent:
if parent.val < val:
return True
if parent.left and parent.left.val < val:
return True
return False
def prev(self) -> int:
if self.now.left:
self.parent.append(self.now)
self.now = self.now.left
while self.now.right:
self.parent.append(self.now)
self.now = self.now.right
return self.now.val
else:
val = self.now.val
while self.parent:
self.now = self.parent.pop()
if self.now.val < val:
return self.now.val
if self.now.left and self.now.left.val < val :
self.parent.append(self.now)
self.now = self.now.left
while self.now.right:
self.parent.append(self.now)
self.now = self.now.right
return self.now.val
