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Description
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.
Credits:
Special thanks to @ts for adding this problem and creating all test cases.
这道题主要就是考二叉树的中序遍历的非递归形式，需要额外定义一个栈来辅助，二叉搜索树的建树规则就是左<根<右，用中序遍历即可从小到大取出所有节点。代码如下：
/**
* 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() {
TreeNode *n = s.top();
s.pop();
int res = n->val;
if (n->right) {
n = n->right;
while (n) {
s.push(n);
n = n->left;
}
}
return res;
}
private:
stack<TreeNode*> s;
};
/**
* Your BSTIterator will be called like this:
* BSTIterator i = BSTIterator(root);
* while (i.hasNext()) cout << i.next();
*/
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