173. 二叉搜索树迭代器

 

实现一个二叉搜索树迭代器。你将使用二叉搜索树的根节点初始化迭代器。

调用 next() 将返回二叉搜索树中的下一个最小的数。

 

示例：

 

 

BSTIterator iterator = new BSTIterator(root);
iterator.next();    // 返回 3
iterator.next();    // 返回 7
iterator.hasNext(); // 返回 true
iterator.next();    // 返回 9
iterator.hasNext(); // 返回 true
iterator.next();    // 返回 15
iterator.hasNext(); // 返回 true
iterator.next();    // 返回 20
iterator.hasNext(); // 返回 false

 

提示：

• next() 和 hasNext() 操作的时间复杂度是 O(1)，并使用 O(h) 内存，其中 h 是树的高度。
• 你可以假设 next() 调用总是有效的，也就是说，当调用 next() 时，BST 中至少存在一个下一个最小的数。

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class BSTIterator:
    def preorderTraversal(self, root):
        if not root:return []
        stack, res = [root], []
        while stack:
            root = stack.pop()
            if root:
                res.append(root.val)
                if root.right:
                    stack.append(root.right)
                if root.left:
                    stack.append(root.left)
        return res

    def __init__(self, root: TreeNode):
        self.nodes=sorted(self.preorderTraversal(root))

    def next(self) -> int:
        """
        @return the next smallest number
        """
        if self.hasNext():
            return self.nodes.pop(0)
        else:
            return None

    def hasNext(self) -> bool:
        """
        @return whether we have a next smallest number
        """
        return len(self.nodes)>0



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

 

 

 

# 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: TreeNode):
        self.stack = []
        while root:
            self.stack.append(root)
            root = root.left

    def next(self) -> int:
        """
        @return the next smallest number
        """
        node = self.stack.pop()
        r = node.right
        while r:
            self.stack.append(r)
            r = r.left
        return node.val

    def hasNext(self) -> bool:
        """
        @return whether we have a next smallest number
        """
        return 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()