LeetCode 솔루션

173. Binary Search Tree Iterator

Jack 쪽지보내기 메일보내기 자기소개 아이디로 검색 전체게시물 작성
작성일 2022.04.19 20:49

6,716 조회
3 댓글
0 추천
0 비추천
목록

글검색

게시글 쓰기

[ Medium ] 173. 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.

Example 1:

Input
["BSTIterator", "next", "next", "hasNext", "next", "hasNext", "next", "hasNext", "next", "hasNext"]
[[[7, 3, 15, null, null, 9, 20]], [], [], [], [], [], [], [], [], []]
Output
[null, 3, 7, true, 9, true, 15, true, 20, false]

Explanation
BSTIterator bSTIterator = new BSTIterator([7, 3, 15, null, null, 9, 20]);
bSTIterator.next();    // return 3
bSTIterator.next();    // return 7
bSTIterator.hasNext(); // return True
bSTIterator.next();    // return 9
bSTIterator.hasNext(); // return True
bSTIterator.next();    // return 15
bSTIterator.hasNext(); // return True
bSTIterator.next();    // return 20
bSTIterator.hasNext(); // return False

Constraints:

The number of nodes in the tree is in the range [1, 105].
0 <= Node.val <= 106
At most 105 calls will be made to hasNext, and next.

Follow up:

Could you implement next() and hasNext() to run in average O(1) time and use O(h) memory, where h is the height of the tree?

#leetcode, #문제풀이, #시즌3, #medium, #facebook, #microsoft, #bloomberg, #google, #amazon, #salesforce, #stack, #tree, #design, #binary search tree, #binary tree, #iterator

링크

https://leetcode.com/problems/binary-search-tree-iterator/ 1639 회 연결

익명
작성일 2022.04.20 08:18

C++
Runtime: 48 ms, faster than 27.69% of C++ online submissions for Binary Search Tree Iterator.
Memory Usage: 24.3 MB, less than 10.53% of C++ online submissions for Binary Search Tree Iterator.

class BSTIterator {
    vector<int> arr;
    int idx;
    
    void buildArr(TreeNode* root) {
        if (root) {
            buildArr(root->left);
            arr.push_back(root->val);
            buildArr(root->right);
        }
    }
    
public:
    BSTIterator(TreeNode* root) {
        buildArr(root);
        idx = 0;
    }
    
    int next() {
        return arr[idx++];
    }
    
    bool hasNext() {
        return idx < arr.size();
    }
};

익명
작성일 2022.04.20 13:01

class BSTIterator:
    def __init__(self, root: Optional[TreeNode]):
        head = root
        def dfs(head):
            return dfs(head.left) + [head.val] + dfs(head.right) if head else []
        self.ans = dfs(root)
        self.iter = 0
        self.lenLst = len(self.ans)
        print(self.ans)
        
    def next(self) -> int:
        if self.iter < self.lenLst:
            temp = self.ans[self.iter]
            self.iter += 1
            return temp
            
    def hasNext(self) -> bool:
        if self.iter < self.lenLst:
            return True
        else:
            return False

익명
작성일 2022.04.20 14:18

Runtime: 31 ms, faster than 76.77% of C++ online submissions for Binary Search Tree Iterator.
Memory Usage: 25.5 MB, less than 6.08% of C++ online submissions for Binary Search Tree Iterator.

class BSTIterator {
private :
    int n=0;
    std::vector<TreeNode*> TreeNodeVec;
public:
    BSTIterator(TreeNode* root) {
        Treesort(root);
    }
    
    void Treesort(TreeNode* p){
            if(p == nullptr){
            return;
        }
        if(p->left != nullptr)
            Treesort(p->left);
        
        TreeNodeVec.push_back(p);
        
        if(p->right != nullptr)
            Treesort(p->right);
    }
    
    
    int next() {
        n++;
        return TreeNodeVec[n-1]->val;
    }
    
    bool hasNext() {
        return TreeNodeVec.size() >= n+1;  
    }
};