题意：实现一个BST的Next（）函数，输出BST里的从小到大的数字。
题解：题目说Next()的时间效率O(1),空间效率O(h),h为树的高度。我们维护一个栈，把前序遍历的左子树的结果存进去。
每次Next取出栈顶元素的时候，再遍历栈顶元素的右子树的前序遍历的左子树部分。
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class BSTIterator {
public:
vector<TreeNode*> stack;
BSTIterator(TreeNode* root) {
if(root!=NULL)
DFS(root);
}
/** @return the next smallest number */
int next() {
TreeNode* term = stack[stack.size()-1];
stack.pop_back();
if(term->right!=NULL)
{
DFS(term->right);
}
return term->val;
}
/** @return whether we have a next smallest number */
bool hasNext() {
if(stack.size()!=0)
return true;
else
return false;
}
void DFS(TreeNode* term)
{
stack.push_back(term);
if(term->left!=NULL)
DFS(term->left);
}
};
/**
* Your BSTIterator object will be instantiated and called as such:
* BSTIterator* obj = new BSTIterator(root);
* int param_1 = obj->next();
* bool param_2 = obj->hasNext();
*/