题目地址:https://leetcode.com/problems/binary-search-tree-iterator/description/

题目描述

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.

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.

Credits: Special thanks to @ts for adding this problem and creating all test cases.

题目大意

实现BST的从大到小依次输出值的操作。实现两个函数,hasNext()next(),操作的时间复杂度是O(1),空间复杂度是O(h)。

解题方法

保存全部节点

一般地,对时间要求比较严格的,我们可以使用空间进行补偿。所以使用一个栈,在初始化的过程中,就使用中序遍历,把BST的中序遍历是有序的这个特点用上。再定义hasnext()和next()就很容易了。

对中序遍历进行了小改进,导致是降序排列的。

但是我们要注意的是,下面的做法的空间复杂度是O(N)的,所以严格来说是不符合题目要求的。

python代码如下:

# 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.inOrder(root)
    
    def inOrder(self, root):
        if not root:
            return
        self.inOrder(root.right)
        self.stack.append(root.val)
        self.inOrder(root.left)
    
    def hasNext(self):
        """
        :rtype: bool
        """
        return len(self.stack) > 0

    def next(self):
        """
        :rtype: int
        """
        return self.stack.pop()

# Your BSTIterator will be called like this:
# i, v = BSTIterator(root), []
# while i.hasNext(): v.append(i.next())

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37

只保留左节点

下面的做法的空间复杂度是O(h),做法是每次保存要遍历的节点的所有左孩子。这样,每次最多也就是H个节点被保存,当遍历了这个节点之后,需要把该节点的右孩子的所有左孩子放到栈里,这就是个中序遍历的过程。

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

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

    def push_left(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()

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41

DDKK.COM 弟弟快看-教程,程序员编程资料站,版权归原作者所有

本文经作者:负雪明烛 授权发布,任何组织或个人未经作者授权不得转发