https://leetcode.com/problems/binary-search-tree-iterator/#/description
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.
Sol:
https://discuss.leetcode.com/topic/6575/my-solutions-in-3-languages-with-stack
I use Stack to store directed left children from root.
When next() be called, I just pop one element and process its right child as new root.
The code is pretty straightforward.
https://discuss.leetcode.com/topic/6629/two-python-solutions-stack-and-generator
# 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 """ # stack stores all left nodes self.stack = [] while root: self.stack.append(root) root = root.left # @return a boolean, whether we have a next smallest number def hasNext(self): """ :rtype: bool """ return len(self.stack) > 0 # @return an integer, the next smallest number def next(self): """ :rtype: int """ # node is the left leaf from bottom node = self.stack.pop() x = node.right while x: self.stack.append(x) x = x.left return node.val # Your BSTIterator will be called like this: # i, v = BSTIterator(root), [] # while i.hasNext(): v.append(i.next())
173. Binary Search Tree Iterator - Unsolved
原文:http://www.cnblogs.com/prmlab/p/7134348.html