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.
基本思路:
事实上就是中序遍历的非递实现。
这next smallest意思就是。从小到大逐个返回。 第一次看到,居然理解反了。还以为是从大到小逐个返回。
/**
* 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) {
s_.push(root);
root = root->left;
}
}
/** @return whether we have a next smallest number */
bool hasNext() {
return !s_.empty();
}
/** @return the next smallest number */
int next() {
auto root = s_.top();
s_.pop();
auto ans = root->val;
root = root->right;
while (root) {
s_.push(root);
root = root->left;
}
return ans;
}
private:
stack<TreeNode *> s_;
};
/**
* Your BSTIterator will be called like this:
* BSTIterator i = BSTIterator(root);
* while (i.hasNext()) cout << i.next();
*/