題目
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.
給一個二叉搜索樹,寫它的迭代器,要求用next()方法返回當前最小的元素,用hasNext()判斷是否還有元素,利用一個棧s保存從根節點到最左的所有元素,根據棧s的空與否判斷hasNext(),每次調用next()返回棧s的棧頂元素的值,同時對該元素從右孩子開始到最左側的進行壓棧,因爲這些是下一次需要的元素。
/**
* 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){//初始化爲從根節點到最左側的所有元素
s.push(root);
root=root->left;
}
}
/** @return whether we have a next smallest number */
bool hasNext() {
return !s.empty();//根據棧是否爲空進行判斷
}
/** @return the next smallest number */
int next() {
TreeNode* temp=s.top();//返回棧頂元素並彈棧
s.pop();
TreeNode* l=temp->right;//從棧頂的右孩子開始,到最左側的所有元素壓棧
while(l!=NULL){
s.push(l);
l=l->left;
}
return temp->val;
}
stack<TreeNode*> s;
};
/**
* Your BSTIterator will be called like this:
* BSTIterator i = BSTIterator(root);
* while (i.hasNext()) cout << i.next();
*/