Leetcode 173. Binary Search Tree Iterator

最新推荐文章于 2025-02-03 15:37:10 发布

xinqrs01

最新推荐文章于 2025-02-03 15:37:10 发布

阅读量172

点赞数

CC 4.0 BY-SA版权

文章标签： tree inorder

本文链接：https://blog.youkuaiyun.com/xinqrs01/article/details/54922310

本文介绍了一种遍历二叉搜索树(BST)的方法，通过迭代器实现平均O(1)时间复杂度的next()和hasNext()操作。利用栈进行中序遍历，确保每次都能返回树中的下一个最小值。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

Note: next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.

s思路：
1. 遍历树，用iterative。in-order遍历的套路。

class BSTIterator {

private:
    stack<TreeNode*> ss;
    TreeNode* cur;
public:
    BSTIterator(TreeNode *root) {
        cur=root;
        while(cur){
            ss.push(cur);
            cur=cur->left;
        }
    }

    /** @return whether we have a next smallest number */
    bool hasNext() {
        return !ss.empty();
    }

    /** @return the next smallest number */
    int next() {
        cur=ss.top();
        ss.pop();
        int res=cur->val;
        cur=cur->right;
        while(cur){
            ss.push(cur);
            cur=cur->left;
        }
        return res;
    }
};

/**
 * Your BSTIterator will be called like this:
 * BSTIterator i = BSTIterator(root);
 * while (i.hasNext()) cout << i.next();
 */