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.

题意

设计二叉搜索树的迭代器

思路1

• 将中序遍历结果存储在一个 vector 中
• 使用 idx 指示当前取到那个位置了

时间复杂度 $O(N)$ 空间复杂度 $O(N)$

代码1

/**
 * 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 {
    vector<int> ans;
    int idx = 0;
public:
    BSTIterator(TreeNode* root) {
        inorder(root);
    }
    
    void inorder(TreeNode* root){
        if(root == NULL) return;
        inorder(root->left);
        ans.push_back(root->val);
        inorder(root->right);
    }
    
    /** @return the next smallest number */
    int next() {
        return ans[idx++];
    }
    
    /** @return whether we have a next smallest number */
    bool hasNext() {
        if(idx < ans.size())
            return true;
        else
            return false;
    }
};

/**
 * 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();
 */

思路2

代码2