173 Binary Search Tree Iterator 二叉搜索树迭代器
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.
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.
题目描述:
实现一个二叉搜索树迭代器。你将使用二叉搜索树的根节点初始化迭代器。
调用 next() 将返回二叉搜索树中的下一个最小的数。
示例 :
BSTIterator iterator = new BSTIterator(root);
iterator.next(); // 返回 3
iterator.next(); // 返回 7
iterator.hasNext(); // 返回 true
iterator.next(); // 返回 9
iterator.hasNext(); // 返回 true
iterator.next(); // 返回 15
iterator.hasNext(); // 返回 true
iterator.next(); // 返回 20
iterator.hasNext(); // 返回 false
提示：
next() 和 hasNext() 操作的时间复杂度是 O(1)，并使用 O(h) 内存，其中 h 是树的高度。
你可以假设 next() 调用总是有效的，也就是说，当调用 next() 时，BST 中至少存在一个下一个最小的数。
思路:
由于需要递增存储二叉树, 采用中序遍历
这里新建一个中序遍历的栈需要时间复杂度 O(n)
由于只有 n个结点, 每次 next()也只要返回 n个结点, 实际上均摊时间复杂度 O(1)
时间复杂度O(1), 空间复杂度O(n)
代码:
C++:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class BSTIterator
{
private:
stack s;
public:
BSTIterator(TreeNode* root)
{
while (root)
{
s.push(root);
root = root -> left;
}
}
/** @return the next smallest number */
int next()
{
TreeNode* cur = s.top();
s.pop();
int result = cur -> val;
cur = cur -> right;
while (cur)
{
s.push(cur);
cur = cur -> left;
}
return result;
}
/** @return whether we have a next smallest number */
bool hasNext()
{
return !s.empty();
}
};
/**
* Your BSTIterator object will be instantiated and called as such:
* BSTIterator* obj = new BSTIterator(root);
* int param_1 = obj->next();
* bool param_2 = obj->hasNext();
*/
Java:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class BSTIterator {
private Stack s;
public BSTIterator(TreeNode root) {
s = new Stack<>();
while (root != null) {
s.push(root);
root = root.left;
}
}
/** @return the next smallest number */
public int next() {
TreeNode temp = s.pop();
int result = temp.val;
temp = temp.right;
while (temp != null) {
s.push(temp);
temp = temp.left;
}
return result;
}
/** @return whether we have a next smallest number */
public boolean hasNext() {
return !s.isEmpty();
}
}
/**
* Your BSTIterator object will be instantiated and called as such:
* BSTIterator obj = new BSTIterator(root);
* int param_1 = obj.next();
* boolean param_2 = obj.hasNext();
*/
Python:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class BSTIterator:
def __init__(self, root: TreeNode):
self.s = []
while root:
self.s.append(root)
root = root.left
def next(self) -> int:
"""
@return the next smallest number
"""
temp = self.s.pop()
result = temp.val
temp = temp.right
while temp:
self.s.append(temp)
temp = temp.left
return result
def hasNext(self) -> bool:
"""
@return whether we have a next smallest number
"""
return len(self.s) != 0
# Your BSTIterator object will be instantiated and called as such:
# obj = BSTIterator(root)
# param_1 = obj.next()
# param_2 = obj.hasNext()