173. 二叉搜索树迭代器
分别结合 栈 和 队列 两种数据结构实现。

class BSTIterator:"""结合栈173. 二叉搜索树迭代器https://leetcode.cn/problems/binary-search-tree-iterator/description/?show=1"""def __init__(self, root: Optional[TreeNode]):self.stk = []self.pushleftBranch(root)# 左侧树枝一撸到底def pushleftBranch(self, p):while p:self.stk.append(p)p = p.leftdef next(self) -> int:p = self.stk.pop()self.pushleftBranch(p.right)return p.valdef hasNext(self) -> bool:return len(self.stk) > 0def peek(self) -> bool:return self.stk[-1].valclass BSTIterator1:"""结合队列173. 二叉搜索树迭代器https://leetcode.cn/problems/binary-search-tree-iterator/description/?show=1"""def __init__(self, root: Optional[TreeNode]):self.deque = collections.deque()self.inOrder(root)def inOrder(self, root):"""二叉树中序遍历:param root::return:"""if not root:returnself.inOrder(root.left)self.deque.append(root.value)self.inOrder(root.right)def next(self) -> int:val = self.deque.popleft()return valdef hasNext(self) -> bool:return len(self.deque) > 0def peek(self) -> bool:return self.deque[0]

BST迭代器应用

1305. 两棵二叉搜索树中的所有元素

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:"""1305. 两棵二叉搜索树中的所有元素"""def getAllElements(self, root1: TreeNode, root2: TreeNode) -> List[int]:t1 = BSTIterator(root1)t2 = BSTIterator(root2)res = []while t1.hasNext() and t2.hasNext():if t1.peek() > t2.peek():res.append(t2.next())else:res.append(t1.next())while t1.hasNext():res.append(t1.next())while t2.hasNext():res.append(t2.next())return resclass BSTIterator:"""结合队列173. 二叉搜索树迭代器https://leetcode.cn/problems/binary-search-tree-iterator/description/?show=1"""def __init__(self, root: Optional[TreeNode]):self.deque = collections.deque()self.inOrder(root)def inOrder(self, root):"""二叉树中序遍历:param root::return:"""if not root:returnself.inOrder(root.left)self.deque.append(root.val)self.inOrder(root.right)def next(self) -> int:val = self.deque.popleft()return valdef hasNext(self) -> bool:return len(self.deque) > 0def peek(self) -> bool:return self.deque[0]