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.
題目題意感覺不好理解,查了下大概是按中序遍歷把節點非遞歸輸出
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class BSTIterator {
public:
stack<TreeNode*>s;
BSTIterator(TreeNode *root) {
while(root != NULL){
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() {
TreeNode * tmp = s.top();
s.pop();
if(tmp->right != NULL){
TreeNode* p = tmp->right;
while(p){
s.push(p);
p = p->left;
}
}
return tmp->val;
}
};
/**
* Your BSTIterator will be called like this:
* BSTIterator i = BSTIterator(root);
* while (i.hasNext()) cout << i.next();
*/