leetcode 第173题 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.

思路：增加一个辅助栈，空间复杂度为O（h）,每次返回的栈顶元素都是二叉搜索树的最小节点。

C++代码实现：

/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class BSTIterator {
private:
stack<TreeNode*> st;
public:
BSTIterator(TreeNode *root) {
findleft(root);
}

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

/** @return the next smallest number */
int next() {
TreeNode *top = st.top();
st.pop();
if(top->right != NULL)
findleft(top->right);
return top->val;
}

void findleft(TreeNode *root){
TreeNode *p = root;
while(p != NULL){
st.push(p);
p = p->left;
}
}
};

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