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算法的重要性，我就不多說了吧，想去大廠，就必須要經(jīng)過基礎(chǔ)知識和業(yè)務(wù)邏輯面試+算法面試。所以，為了提高大家的算法能力，這個公眾號后續(xù)每天帶大家做一道算法題，題目就從LeetCode上面選！

今天和大家聊的問題叫做?二叉搜索樹迭代器??，我們先來看題面：

https://leetcode-cn.com/problems/binary-search-tree-iterator/

Implement the BSTIterator class that represents an iterator over the in-order traversal of a binary search tree (BST):

BSTIterator(TreeNode root) Initializes an object of the BSTIterator class. The root of the BST is given as part of the constructor. The pointer should be initialized to a non-existent number smaller than any element in the BST.

boolean hasNext() Returns true if there exists a number in the traversal to the right of the pointer, otherwise returns false.

int next() Moves the pointer to the right, then returns the number at the pointer.

Notice that by initializing the pointer to a non-existent smallest number, the first call to next() will return the smallest element in the BST.

You may assume that next() calls will always be valid. That is, there will be at least a next number in the in-order traversal when next() is called.

題意

實現(xiàn)一個二叉搜索樹迭代器。你將使用二叉搜索樹的根節(jié)點初始化迭代器。

調(diào)用 next() 將返回二叉搜索樹中的下一個最小的數(shù)。

樣例

解題

https://codingchaozhang.blog.csdn.net/article/details/109497176

思路一：對其進(jìn)行遍歷，存儲?對其進(jìn)行用棧來存儲

class?TreeNode{
????int?val;
????TreeNode left;
????TreeNode right;
????TreeNode(int?x){
??????val = x;
????}
??}
??
??class?BSTIterator?{
??????int?index;
??????ArrayList list;

??????public?BSTIterator(TreeNode root)?{
??????????this.list?= new?ArrayList<>();
??????????this.index = -1;
??????????this.inorder(root);
??????}
??????// 先進(jìn)行中序遍歷
??????private?void?inorder(TreeNode root){
??????????if(root==null){
??????????????return;
??????????}
??????????inorder(root.left);
??????????list.add(root.val);
??????????inorder(root.right);
??????}
??????
??????/** @return the next smallest number */
??????public?int?next()?{
??????????return?list.get(++index);
??????}
??????
??????/** @return whether we have a next smallest number */
??????public?boolean hasNext()?{
??????????return?this.index+1?< this.list.size();
??????}
??}

思路二：對其進(jìn)行用棧來存儲

class?BSTIterator?{

??????// 棧
??????Stack stack;
??????public?BSTIterator(TreeNode root)?{
??????????// 初始化
??????????stack?= new?Stack<>();
??????????this.pre(root);
??????}
??????// 先存儲一部分值
??????private?void?pre(TreeNode root){
??????????while(root!=null){
??????????????stack.push(root);
??????????????root = root.left;
??????????}
??????}
??????/** @return the next smallest number */
??????public?int?next()?{
??????????TreeNode temp = this.stack.pop();

??????????if(temp.right!=null){
??????????????this.pre(temp.right);
??????????}

??????????return?temp.val;
??????}
??????
??????/** @return whether we have a next smallest number */
??????public?boolean hasNext()?{
??????????return?this.stack.size()>0;
??????}
??}