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题目
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.
Credits:
Special thanks to @ts for adding this problem and creating all test cases.
实现思路
要找到二叉搜索树中的最小节点,应该从根节点递归遍历左节点,直到遍历的节点没有左节点,那么该节点就是二叉树中的最小节点。现在已经有二叉搜索树中没有访问过的最小节点了,那么当访问了该节点后,剩余没有访问的树中最小的节点在哪里呢?如果该节点有右子树,那么在它的右子树中(又回到了找一棵二叉搜索树的最小元素,不过这棵二叉搜索树变小了);如果没有右子树,那么就是它的父节点。
为了能够快速定位到父节点,我们可以用栈将遍历路径暂存起来,当进行next()
操作时,我们弹出栈顶元素并进行访问,如果它有右子树的话就遍历它的右子树;如果没有右子树,当下次出栈操作时就是访问当前节点的父节点了。
实现代码
# 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._pushLeft(root)
def hasNext(self):
"""
:rtype: bool
"""
return self.stack
def next(self):
"""
:rtype: int
"""
node = self.stack.pop()
self._pushLeft(node.right)
return node.val
def _pushLeft(self,node):
while node:
self.stack.append(node)
node = node.left
# Your BSTIterator will be called like this:
# i, v = BSTIterator(root), []
# while i.hasNext(): v.append(i.next())