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.

采用中序遍历将节点压入栈中，由于要求存储空间为O（h），因此不能在一开始将所有节点全部压入，只是压入最左边一列。当取出一个节点时，压入其右子节点的所有左节点。

/**
 * 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*> stk;
    int minres;
public:
    BSTIterator(TreeNode *root) {
        while(root)
        {
            stk.push(root);
            root=root->left;
        }
    }

    /** @return whether we have a next smallest number */
    bool hasNext() {
        if(stk.empty())
            return false;
        TreeNode* top;
        top=stk.top();
        minres=top->val;
        stk.pop();
        TreeNode* cur=top->right;
        if(cur)
        {
            stk.push(cur);
            cur=cur->left;
            while(cur)
            {
                stk.push(cur);
                cur=cur->left;
            }
        }
        return true;
    }

    /** @return the next smallest number */
    int next() {
        return minres;
    }
};

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