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实现一个二叉搜索树迭代器类BSTIterator ，表示一个按中序遍历二叉搜索树（BST）的迭代器：
BSTIterator(TreeNode root)初始化BSTIterator类的一个对象。BST 的根节点root会作为构造函数的一部分给出。指针应初始化为一个不存在于 BST 中的数字，且该数字小于 BST 中的任何元素。boolean hasNext()如果向指针右侧遍历存在数字，则返回true；否则返回false。int next()将指针向右移动，然后返回指针处的数字。
注意，指针初始化为一个不存在于 BST 中的数字，所以对 next() 的首次调用将返回 BST 中的最小元素。
你可以假设 next() 调用总是有效的，也就是说，当调用 next() 时，BST 的中序遍历中至少存在一个下一个数字。
示例：
输入 ["BSTIterator", "next", "next", "hasNext", "next", "hasNext", "next", "hasNext", "next", "hasNext"] [[[7, 3, 15, null, null, 9, 20]], [], [], [], [], [], [], [], [], []] 输出 [null, 3, 7, true, 9, true, 15, true, 20, false]解释 BSTIterator bSTIterator = new BSTIterator([7, 3, 15, null, null, 9, 20]); bSTIterator.next(); // 返回 3 bSTIterator.next(); // 返回 7 bSTIterator.hasNext(); // 返回 True bSTIterator.next(); // 返回 9 bSTIterator.hasNext(); // 返回 True bSTIterator.next(); // 返回 15 bSTIterator.hasNext(); // 返回 True bSTIterator.next(); // 返回 20 bSTIterator.hasNext(); // 返回 False
提示：
- 树中节点的数目在范围
[1, 105]内 0 <= Node.val <= 106- 最多调用
105次hasNext和next操作
进阶：
- 你可以设计一个满足下述条件的解决方案吗？
next()和hasNext()操作均摊时间复杂度为O(1)，并使用O(h)内存。其中h是树的高度。
初始化数据时，递归中序遍历，将二叉搜索树每个结点的值保存在列表 vals 中。用 cur 指针记录外部即将遍历的位置，初始化为 0。
调用 next() 时，返回 vals[cur]，同时 cur 指针自增。调用 hasNext() 时，判断 cur 指针是否已经达到 len(vals) 个数，若是，说明已经遍历结束，返回 false，否则返回 true。
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class BSTIterator:
def __init__(self, root: TreeNode):
def inorder(root):
if root:
inorder(root.left)
self.vals.append(root.val)
inorder(root.right)
self.cur = 0
self.vals = []
inorder(root)
def next(self) -> int:
res = self.vals[self.cur]
self.cur += 1
return res
def hasNext(self) -> bool:
return self.cur < len(self.vals)
# Your BSTIterator object will be instantiated and called as such:
# obj = BSTIterator(root)
# param_1 = obj.next()
# param_2 = obj.hasNext()/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class BSTIterator {
private int cur = 0;
private List<Integer> vals = new ArrayList<>();
public BSTIterator(TreeNode root) {
inorder(root);
}
public int next() {
return vals.get(cur++);
}
public boolean hasNext() {
return cur < vals.size();
}
private void inorder(TreeNode root) {
if (root != null) {
inorder(root.left);
vals.add(root.val);
inorder(root.right);
}
}
}
/**
* 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();
*//**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class BSTIterator {
public:
vector<int> vals;
int cur;
BSTIterator(TreeNode* root) {
cur = 0;
inorder(root);
}
int next() {
return vals[cur++];
}
bool hasNext() {
return cur < vals.size();
}
void inorder(TreeNode* root) {
if (root) {
inorder(root->left);
vals.push_back(root->val);
inorder(root->right);
}
}
};
/**
* 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();
*//**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
type BSTIterator struct {
stack []*TreeNode
}
func Constructor(root *TreeNode) BSTIterator {
var stack []*TreeNode
for ; root != nil; root = root.Left {
stack = append(stack, root)
}
return BSTIterator{
stack: stack,
}
}
func (this *BSTIterator) Next() int {
cur := this.stack[len(this.stack)-1]
this.stack = this.stack[:len(this.stack)-1]
for node := cur.Right; node != nil; node = node.Left {
this.stack = append(this.stack, node)
}
return cur.Val
}
func (this *BSTIterator) HasNext() bool {
return len(this.stack) > 0
}
/**
* Your BSTIterator object will be instantiated and called as such:
* obj := Constructor(root);
* param_1 := obj.Next();
* param_2 := obj.HasNext();
*//**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
class BSTIterator {
private data: number[];
private index: number;
constructor(root: TreeNode | null) {
this.index = 0;
this.data = [];
const dfs = (root: TreeNode | null) => {
if (root == null) {
return;
}
const { val, left, right } = root;
dfs(left);
this.data.push(val);
dfs(right);
};
dfs(root);
}
next(): number {
return this.data[this.index++];
}
hasNext(): boolean {
return this.index < this.data.length;
}
}
/**
* Your BSTIterator object will be instantiated and called as such:
* var obj = new BSTIterator(root)
* var param_1 = obj.next()
* var param_2 = obj.hasNext()
*/// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
// pub val: i32,
// pub left: Option<Rc<RefCell<TreeNode>>>,
// pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
// #[inline]
// pub fn new(val: i32) -> Self {
// TreeNode {
// val,
// left: None,
// right: None
// }
// }
// }
struct BSTIterator {
vals: Vec<i32>,
index: usize,
}
use std::rc::Rc;
use std::cell::RefCell;
/**
* `&self` means the method takes an immutable reference.
* If you need a mutable reference, change it to `&mut self` instead.
*/
impl BSTIterator {
fn inorder(root: &Option<Rc<RefCell<TreeNode>>>, res: &mut Vec<i32>) {
if let Some(node) = root {
let node = node.as_ref().borrow();
Self::inorder(&node.left, res);
res.push(node.val);
Self::inorder(&node.right, res);
}
}
fn new(root: Option<Rc<RefCell<TreeNode>>>) -> Self {
let mut vals = vec![];
Self::inorder(&root, &mut vals);
BSTIterator {
vals,
index: 0,
}
}
fn next(&mut self) -> i32 {
self.index += 1;
self.vals[self.index - 1]
}
fn has_next(&self) -> bool {
self.index != self.vals.len()
}
}/**
* Your BSTIterator object will be instantiated and called as such:
* let obj = BSTIterator::new(root);
* let ret_1: i32 = obj.next();
* let ret_2: bool = obj.has_next();
*//**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
*/
var BSTIterator = function (root) {
this.stack = [];
for (; root != null; root = root.left) {
this.stack.push(root);
}
};
/**
* @return {number}
*/
BSTIterator.prototype.next = function () {
let cur = this.stack.pop();
let node = cur.right;
for (; node != null; node = node.left) {
this.stack.push(node);
}
return cur.val;
};
/**
* @return {boolean}
*/
BSTIterator.prototype.hasNext = function () {
return this.stack.length > 0;
};
/**
* Your BSTIterator object will be instantiated and called as such:
* var obj = new BSTIterator(root)
* var param_1 = obj.next()
* var param_2 = obj.hasNext()
*/初始化时，从根节点一路遍历所有左子节点，压入栈 stack 中。
调用 next()时，弹出栈顶元素 cur，获取 cur 的右子节点 node，若 node 不为空，一直循环压入左节点。最后返回 cur.val 即可。调用 hasNext() 时，判断 stack 是否为空，空则表示迭代结束。
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class BSTIterator:
def __init__(self, root: TreeNode):
self.stack = []
while root:
self.stack.append(root)
root = root.left
def next(self) -> int:
cur = self.stack.pop()
node = cur.right
while node:
self.stack.append(node)
node = node.left
return cur.val
def hasNext(self) -> bool:
return len(self.stack) > 0
# Your BSTIterator object will be instantiated and called as such:
# obj = BSTIterator(root)
# param_1 = obj.next()
# param_2 = obj.hasNext()/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class BSTIterator {
private Deque<TreeNode> stack = new LinkedList<>();
public BSTIterator(TreeNode root) {
for (; root != null; root = root.left) {
stack.offerLast(root);
}
}
public int next() {
TreeNode cur = stack.pollLast();
for (TreeNode node = cur.right; node != null; node = node.left) {
stack.offerLast(node);
}
return cur.val;
}
public boolean hasNext() {
return !stack.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();
*//**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class BSTIterator {
public:
stack<TreeNode*> stack;
BSTIterator(TreeNode* root) {
for (; root != nullptr; root = root->left) {
stack.push(root);
}
}
int next() {
TreeNode* cur = stack.top();
stack.pop();
TreeNode* node = cur->right;
for (; node != nullptr; node = node->left) {
stack.push(node);
}
return cur->val;
}
bool hasNext() {
return !stack.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();
*//**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
class BSTIterator {
private stack: TreeNode[];
constructor(root: TreeNode | null) {
this.stack = [];
const dfs = (root: TreeNode | null) => {
if (root == null) {
return;
}
this.stack.push(root);
dfs(root.left);
};
dfs(root);
}
next(): number {
const { val, right } = this.stack.pop();
if (right) {
let cur = right;
while (cur != null) {
this.stack.push(cur);
cur = cur.left;
}
}
return val;
}
hasNext(): boolean {
return this.stack.length !== 0;
}
}
/**
* Your BSTIterator object will be instantiated and called as such:
* var obj = new BSTIterator(root)
* var param_1 = obj.next()
* var param_2 = obj.hasNext()
*/// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
// pub val: i32,
// pub left: Option<Rc<RefCell<TreeNode>>>,
// pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
// #[inline]
// pub fn new(val: i32) -> Self {
// TreeNode {
// val,
// left: None,
// right: None
// }
// }
// }
struct BSTIterator {
stack: Vec<Option<Rc<RefCell<TreeNode>>>>,
}
use std::rc::Rc;
use std::cell::RefCell;
/**
* `&self` means the method takes an immutable reference.
* If you need a mutable reference, change it to `&mut self` instead.
*/
impl BSTIterator {
fn dfs(
mut root: Option<Rc<RefCell<TreeNode>>>,
stack: &mut Vec<Option<Rc<RefCell<TreeNode>>>>
) {
if root.is_some() {
let left = root.as_mut().unwrap().borrow_mut().left.take();
stack.push(root);
Self::dfs(left, stack);
}
}
fn new(root: Option<Rc<RefCell<TreeNode>>>) -> Self {
let mut stack = vec![];
Self::dfs(root, &mut stack);
BSTIterator { stack }
}
fn next(&mut self) -> i32 {
let node = self.stack.pop().unwrap().unwrap();
let mut node = node.borrow_mut();
if node.right.is_some() {
Self::dfs(node.right.take(), &mut self.stack);
}
node.val
}
fn has_next(&self) -> bool {
self.stack.len() != 0
}
}/**
* Your BSTIterator object will be instantiated and called as such:
* let obj = BSTIterator::new(root);
* let ret_1: i32 = obj.next();
* let ret_2: bool = obj.has_next();
*/