173. Binary Search Tree Iterator

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本文介绍了一种二叉搜索树迭代器的实现方法，该迭代器可以在平均O(1)的时间复杂度内返回二叉搜索树中的下一个最小元素，并使用了O(h)的空间复杂度，其中h为树的高度。通过结合栈和树节点的数据结构，利用二叉搜索树的中序遍历特性实现了迭代器的功能。

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Binary Search Tree Iterator
- Description
- Implementation

173. Binary Search Tree Iterator

Description

mplement 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.

Implementation

Data structure is combined with Stack and a TreeNode, the tree node has left and right nodes, and the stack is used to store the next number for smallest number.

From the problem we can know, we can know that the this is a binary search problem with in-order search. So in the hasNode I add the left-part value into it and in the output part with pop node, we will set the right node for current node to search the right part.

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class BSTIterator {
public:
    TreeNode* cur;
    stack<TreeNode*> stk;

    BSTIterator(TreeNode *root) {
        this->cur = root;
    }

    void addNode(TreeNode* cur) {
        while(this->cur != NULL) {
            stk.push(this->cur);
            this->cur = this->cur->left;
        }
    }

    /** @return whether we have a next smallest number */
    bool hasNext() {
        if(this->cur != NULL) {
            addNode(this->cur);
            return true;
        }
        else if(stk.size() > 0) 
            return true;
        return false;    
    }

    /** @return the next smallest number */
    int next() {
        TreeNode* tp = stk.top();
        stk.pop();
        cur = tp->right;
        return tp->val;
    }
};