题目:原题链接(中等)
标签:树、二叉树、二叉树-遍历、二叉搜索树、栈、设计
相关题目:0094(实际上就是二叉搜索树的中序遍历)
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
---|---|---|---|
Ans 1 (Python) | 116ms (40.10%) | ||
Ans 2 (Python) | 104ms (81.09%) | ||
Ans 3 (Python) | 96ms (96.38%) |
LeetCode的Python执行用时随缘,只要时间复杂度没有明显差异,执行用时一般都在同一个量级,仅作参考意义。
解法一(初始化完成中序遍历):
class BSTIterator:
def __init__(self, root: TreeNode):
stack = [root]
self.ans = []
while stack:
node = stack[-1]
if node:
stack.append(node.left)
else:
stack.pop()
if not stack:
break
now = stack.pop()
self.ans.append(now.val)
stack.append(now.right)
def next(self) -> int:
return self.ans.pop(0)
def hasNext(self) -> bool:
return len(self.ans) > 0
解法二(解法一的优化):
class BSTIterator:
def __init__(self, root: TreeNode):
self.stack = [root]
def next(self) -> int:
while self.stack:
node = self.stack[-1]
if node:
self.stack.append(node.left)
else:
self.stack.pop()
if not self.stack:
break
now = self.stack.pop()
self.stack.append(now.right)
return now.val
def hasNext(self) -> bool:
return len(self.stack) > 0 and self.stack[0] is not None
解法三:
class BSTIterator:
def __init__(self, root: TreeNode):
self.stack = []
while root:
self.stack.append(root)
root = root.left
def next(self) -> int:
node = self.stack.pop()
ans = node.val
node = node.right
while node:
self.stack.append(node)
node = node.left
return ans
def hasNext(self) -> bool:
return len(self.stack) > 0 and self.stack[0] is not None