10
04/2015
[LeetCode] Binary Search Tree Iterator
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.
解题思路：
运用栈的技术，将当前未被访问的最左边的一条路径入栈，每次取值的时候将栈顶元素弹出，并将栈顶元素的右子树的最左边一条路径入栈。hasNext()只需要看栈是否为空即可。但是有个问题，似乎next()函数的时间复杂度并不是O(1)，因为我们要维护栈，因此是O(nlgn)。不知有什么更为好的方法没有。我还犯了一个小错误，就是stack<TreeNode*> stack。不能取名叫stack呀。
/**
* 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) {
pushLeftChildIntoStack(root);
}
/** @return whether we have a next smallest number */
bool hasNext() {
return !s.empty();
}
/** @return the next smallest number */
int next() {
TreeNode* node=s.top();
s.pop();
pushLeftChildIntoStack(node->right);
return node->val;
}
private:
stack<TreeNode*> s;
void pushLeftChildIntoStack(TreeNode* node){
while(node!=NULL){
s.push(node);
node=node->left;
}
}
};
/**
* Your BSTIterator will be called like this:
* BSTIterator i = BSTIterator(root);
* while (i.hasNext()) cout << i.next();
*/
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