173. Binary Search Tree Iterator

tree stack

Implement the BSTIterator class that represents an iterator over the in-order traversal of a binary search tree (BST):

• BSTIterator(TreeNode root) Initializes an object of the BSTIterator class. The root of the BST is given as part of the constructor. The pointer should be initialized to a non-existent number smaller than any element in the BST.

• boolean hasNext() Returns true if there exists a number in the traversal to the right of the pointer, otherwise returns false.

• int next() Moves the pointer to the right, then returns the number at the pointer.

Notice that by initializing the pointer to a non-existent smallest number, the first call to next() will return the smallest element in the BST.

You may assume that next() calls will always be valid. That is, there will be at least a next number in the in-order traversal when next() is called.

Example 1:

Input
["BSTIterator", "next", "next", "hasNext", "next", "hasNext", "next", "hasNext", "next", "hasNext"]
[[[7, 3, 15, null, null, 9, 20]], [], [], [], [], [], [], [], [], []]
Output
[null, 3, 7, true, 9, true, 15, true, 20, false]

Explanation
BSTIterator bSTIterator = new BSTIterator([7, 3, 15, null, null, 9, 20]);
bSTIterator.next();    // return 3
bSTIterator.next();    // return 7
bSTIterator.hasNext(); // return True
bSTIterator.next();    // return 9
bSTIterator.hasNext(); // return True
bSTIterator.next();    // return 15
bSTIterator.hasNext(); // return True
bSTIterator.next();    // return 20
bSTIterator.hasNext(); // return False

Constraints:

• The number of nodes in the tree is in the range [1, 105].

• 0 <= Node.val <= 106

• At most 105 calls will be made to hasNext, and next.

Follow up:

• Could you implement next() and hasNext() to run in average O(1) time and use O(h) memory, where h is the height of the tree?

分析

✅思路（Lazy 中序遍历）

中序遍历 BST 的顺序是：左 → 根 → 右。 我们用一个栈模拟这个过程，并只存一条路径：

✅初始化时：

• 一路往左压栈，把最小的元素放到栈顶

✅每次调用 next()：

• 弹出栈顶元素（当前最小）

• 然后如果这个节点有右子树，把它的右子树的左边一条路径压栈

# 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 BSTIterator:

    def __init__(self, root: Optional[TreeNode]):
        self.stack = []
        self.current = root

    def next(self) -> int:
        while self.current:
            self.stack.append(self.current)
            self.current = self.current.left
        self.current = self.stack.pop()
        val = self.current.val
        self.current = self.current.right
        return val

    def hasNext(self) -> bool:
        return self.current is not None or len(self.stack) > 0

# Your BSTIterator object will be instantiated and called as such:
# obj = BSTIterator(root)
# param_1 = obj.next()
# param_2 = obj.hasNext()