LeetCode: Binary Search Tree Iterator Posted on January 13, 2018July 26, 2020 by braindenny Binary Search Tree Iterator Similar Problems: CheatSheet: Leetcode For Code Interview CheatSheet: Common Code Problems & Follow-ups Tag: #oodesign Implement 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. Note: next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree. Github: code.dennyzhang.com Credits To: leetcode.com Leave me comments, if you have better ways to solve. ## https://code.dennyzhang.com/binary-search-tree-iterator ## Basic Ideas: In-order traversal. Use a stack ## Store directed left children from root. ## When next() be called, pop one element and process its right child as new root. ## Complexity: ## # 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 = [] p = root while p: self.stack.append(p) p = p.left def hasNext(self): """ :rtype: bool """ return len(self.stack) > 0 def next(self): """ :rtype: int """ node = self.stack.pop() p = node.right while p: self.stack.append(p) p = p.left return node.val # Your BSTIterator will be called like this: # i, v = BSTIterator(root), [] # while i.hasNext(): v.append(i.next()) Post Views: 2