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  for adding this problem and creating all test cases.

 

这道题主要就是考二叉树的中序遍历的非递归形式，需要额外定义一个栈来辅助，二叉搜索树的建树规则就是左<根<右，用中序遍历即可从小到大取出所有节点。代码如下：

 

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