[LeetCode] 173. Binary Search Tree Iterator_Medium_tag: Binary Search Tree

Posted 2020-11-23

mplement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST.

Calling next() will return the next smallest number in the BST.

 

Example:

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

 

Note:

• next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.
• You may assume that next() call will always be valid, that is, there will be at least a next smallest number in the BST when next() is called.

 

因为是Binary Search Tree, 所以要用inorder travesal，所以可以在初始化的时候将tree 转变为stack，并且用一个指针来指向现在的node所在的位置，如果超过stack，那么就没有next。

 

T: O(n), S: O(n)

Code：

# 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 inOrder(self, root):
        if not root: return 
        self.inOrder(root.left)
        self.stack.append(root.val)
        self.inOrder(root.right)

    def __init__(self, root):
        """
        :type root: TreeNode
        """
        self.stack = []
        self.point = 0
        self.inOrder(root)
        

    def next(self):
        """
        @return the next smallest number
        :rtype: int
        """
        if self.hasNext():
            nextNum = self.stack[self.point]
            self.point += 1 
            return nextNum
        

    def hasNext(self):
        """
        @return whether we have a next smallest number
        :rtype: bool
        """
        if not self.stack or self.point >= len(self.stack):
            return False
        return True
        


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