LeetCode---Binary Search Tree Iterator ---C++

问题描述：

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.
对BST，实现一个迭代器。
思路：对于BST，其中序遍历为排序的。中序遍历的非递归方法，就用到了一个栈。所以该题的思路就是用一个栈来实现。因此BST的迭代器类，有一个私有的stack，在构造函数中把BST中最左边的节点入栈。然后hasnext（）函数返回true当栈中还有元素时。next，出栈，返回值为该节点的val，然后判断是否有右节点，若有，则把该节点及其所有左节点入栈。

/**
* 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 != NULL )
{
S.push(root);
root = root->left;
}
}

/** @return whether we have a next smallest number */
bool hasNext() {
if(S.size() > 0)
return true;
return false;
}

/** @return the next smallest number */
int next() {
int result = 0;
if(!S.empty())
{
TreeNode* node = S.top();
result = node->val;
S.pop();
node = node->right;
while(node != NULL)
{
S.push(node);
node = node->left;
}

}
return result;
}
private:
std::stack<TreeNode*> S;
};

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