LeetCode[Tree]: Binary Search Tree Iterator，leetcodeiterator

LeetCode[Tree]: Binary Search Tree Iterator，leetcodeiterator

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.

參考：https://oj.leetcode.com/discuss/20001/my-solutions-in-3-languages-with-stack

解題思路：用一個stack儲存從根節點開始的所有左孩子，每次調用next()就從stack裡面pop一個元素，並將以這個節點的右孩子為根節點的子樹重複同樣的過程。

這個演算法滿足O(h)的空間複雜度，hasNext()的時間複雜度滿足O(1)，儘管next()的時間複雜度為O(h)，但是這個演算法的處理過程仍然值得學習。

C++代碼實現如下：

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class BSTIterator {private:    stack<TreeNode *> nodeStack;    void pushAll(TreeNode *root) {        for (TreeNode *node = root; node != nullptr; node = node->left) {            nodeStack.push(node);        }    }public:    BSTIterator(TreeNode *root) {        pushAll(root);    }    /** @return whether we have a next smallest number */    bool hasNext() {        return !nodeStack.empty();    }    /** @return the next smallest number */    int next() {        TreeNode *node = nodeStack.top();        nodeStack.pop();        pushAll(node->right);        return node->val;    }};/** * Your BSTIterator will be called like this: * BSTIterator i = BSTIterator(root); * while (i.hasNext()) cout << i.next(); */

該演算法的時間效能表現如下：