Leetcode 173.二叉搜索树迭代器

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173. 二叉搜索树迭代器

题目

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。

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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

方法

方法1：非递归中序遍历

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

class BSTIterator(object):

    def __init__(self, root):
        """
        :type root: TreeNode
        """
        self.stack=[]
        while root:
            self.stack.append(root)
            root=root.left


    def next(self):
        """
        @return the next smallest number
        :rtype: int
        """
        t=self.stack.pop()
        x=t.right
        while x:
            self.stack.append(x)
            x=x.left
        return t.val


    def hasNext(self):
        """
        @return whether we have a next smallest number
        :rtype: bool
        """
        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()
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# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class BSTIterator(object):

    def __init__(self, root):
        """
        :type root: TreeNode
        """
        self.stack=[]
        self.LeftToStack(root)

    def next(self):
        """
        @return the next smallest number
        :rtype: int
        """
        node=self.stack.pop()
        self.LeftToStack(node.right)
        return node.val

    def hasNext(self):
        """
        @return whether we have a next smallest number
        :rtype: bool
        """
        return bool(self.stack)

    def LeftToStack(self,root):
        while root:
            self.stack.append(root)
            root=root.left

# Your BSTIterator object will be instantiated and called as such:
# obj = BSTIterator(root)
# param_1 = obj.next()
# param_2 = obj.hasNext()
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# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class BSTIterator(object):

    def __init__(self, root):
        """
        :type root: TreeNode
        """
        self.root=root
        self.stack=[]


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


    def hasNext(self):
        """
        @return whether we have a next smallest number
        :rtype: bool
        """
        return bool(self.stack) or bool(self.root)



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