LeetCode No.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.

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題目鏈接：https://leetcode.com/problems/binary-search-tree-iterator/

題目大意：使用 O(1) 的時間複雜度和 O(h) 的空間複雜度實現二叉搜索樹的判斷是否有下一個值hasNext()和輸出下一個值next()的操作。

思路：模擬二叉搜索樹的非遞歸版中序遍歷。

參考代碼：

/**
 * 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) {
        node.clear() ;
        cur = root ;
    }
    
    /** @return whether we have a next smallest number */
    bool hasNext() {
        if ( cur || ! node.empty() )
        {
            while ( cur )
            {
                node.push_back ( cur ) ;
                cur = cur -> left ;
            }
            return true ;
        }
        return false ;
    }

    /** @return the next smallest number */
    int next() {
        cur = node.back() ;
        node.pop_back() ;
        int ans = cur -> val ;
        cur = cur -> right ;
        return ans ;
    }
private :
    vector <TreeNode*> node ;
    TreeNode* cur ;
};

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