173. 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.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

 

开始主要是不知道这道题想表达什么意思，知道他想表达什么意思之后就很简单了。

思路：找最小值，可以参考中序遍历(迭代方法)，借助栈！每弹出一个元素，才增加栈中元素，不用马上遍历整颗树!

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class BSTIterator {
private:
    stack<TreeNode*> stk;
public:
    BSTIterator(TreeNode *root) {
        while(root)
        {
            stk.push(root);
            root=root->left;
        }
    }

    /** @return whether we have a next smallest number */
    bool hasNext() {
        return !stk.empty();
    }

    /** @return the next smallest number */
    int next() {
        TreeNode*temp=stk.top();
        stk.pop();
        int res=temp->val;
        temp=temp->right;
        while(temp)
        {
            stk.push(temp);
            temp=temp->left;
        }
        return res;
    }
};

/**
 * Your BSTIterator will be called like this:
 * BSTIterator i = BSTIterator(root);
 * while (i.hasNext()) cout << i.next();
 */