Binary Search Tree Iterator

Binary Search Tree Iterator

Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST.

Callingnext()Will return the next smallest number in the BST.

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

Credits:
Special thanks to @ ts for adding this problem and creating all test cases.

I initially thought that the point of consideration for this question was to change a BST to a double-stranded table through the middle-order traversal, so I used the following code,

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class BSTIterator {private:    TreeNode *it;    TreeNode *last;    void toList(TreeNode *root){        if(!root)             return;        //save        TreeNode *cur=root;                if(cur->left)            toList(cur->left);                    cur->left=last;        if(last)            last->right=cur;        last=cur;                if(cur->right)            toList(cur->right);    }public:    BSTIterator(TreeNode *root) {        last=NULL;        toList(root);                it=last;        while(it&&it->left){            it=it->left;        }    }    /** @return whether we have a next smallest number */    bool hasNext() {        return (it!=NULL);    }    /** @return the next smallest number */    int next() {        int val= it->val;        it=it->right;        return val;    }};/** * Your BSTIterator will be called like this: * BSTIterator i = BSTIterator(root); * while (i.hasNext()) cout << i.next(); */

The result is miserable ......

Submission Result: Runtime ErrorMore Details

Last executed input:

{2, 1}

Probably because I changed the structure of the BST tree ...... I searched from the Internet and found that they all use stack and sequential traversal ...... Okay, follow suit

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class BSTIterator {private:    stack
 
   s;    void toStack(TreeNode *root){        if(root){            TreeNode *p=root;            while(p){                s.push(p);                p=p->left;            }        }    }public:    BSTIterator(TreeNode *root) {        toStack(root);    }        /** @return whether we have a next smallest number */    bool hasNext() {        return !s.empty();    }    /** @return the next smallest number */    int next() {        TreeNode* n= s.top();        s.pop();        toStack(n->right);        return n->val;    }};/** * Your BSTIterator will be called like this: * BSTIterator i = BSTIterator(root); * while (i.hasNext()) cout << i.next(); */