leetcode之Binary Search Tree Iterator

题目：

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.

解答：

由于BST的中序遍历就是从小到大的顺序， 所以直接进行中序遍历即可，为了满足复杂度要求，需要将递归改为迭代过程，用栈实现

/**
* 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) {
if(!root)
return;
while(root)
{
st.push(root);
root = root->left;
}
}

/** @return whether we have a next smallest number */
bool hasNext() {
return !st.empty();
}

/** @return the next smallest number */
int next() {
if(BSTIterator :: hasNext())
{
int res = st.top()->val;
TreeNode *tmp = st.top();
st.pop();
if(tmp->right)
{
tmp = tmp->right;
while(tmp)
{
st.push(tmp);
tmp = tmp->left;
}
}
return res;
}
}
private:
stack<TreeNode*> st;
};

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