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.

思路

整体思路就是对BST进行前序遍历即可。此题重点在于设计和实现，我们在前序遍历的时候利用栈来进行，避免采用递归的方式。

需要注意利用栈的时候，C++必须利用top()函数来取栈顶的值，而不能用pop()。一定要注意pop()所在的位置。也一定要注意是否会有内存的溢出。 

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/**
 * 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<TreeNode*> DFS;
public:
    BSTIterator(TreeNode *root) {
        while(root != NULL){
            DFS.push(root);
            root = root->left;
        }
    }
    /** @return whether we have a next smallest number */
    bool hasNext() {
        if (DFS.empty()) return false;
        else return true;
    }
    /** @return the next smallest number */
    int next() {
        TreeNode *cur = DFS.top();
        DFS.pop();
        if(cur != NULL) 
        {
            if(cur->right != NULL){
                TreeNode *tmp = cur->right;
                while(tmp != NULL){
                    DFS.push(tmp);
                    tmp = tmp->left;
                }
            }
            return cur->val;
        }
        else return 0;
    }
};
/**
 * Your BSTIterator will be called like this:
 * BSTIterator i = BSTIterator(root);
 * while (i.hasNext()) cout << i.next();
 */