# 173. 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.

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.

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

``````/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class BSTIterator {
public:
BSTIterator(TreeNode *root) {
while (root) {
s.push(root);
root = root->left;
}
}

/** @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();
int res = n->val;
if (n->right) {
n = n->right;
while (n) {
s.push(n);
n = n->left;
}
}
return res;
}
private:
stack<TreeNode*> s;
};

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

LeetCode All in One 题目讲解汇总(持续更新中…)

 微信打赏 Venmo 打赏
（欢迎加入博主的知识星球，博主将及时答疑解惑，并分享刷题经验与总结，试运营期间前五十位可享受半价优惠～）

×

Help us with donation