LeetCode[Tree]: 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.

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

解题思路：用一个stack保存从根节点开始的所有左孩子，每次调用next()就从stack里面pop一个元素，并将以这个节点的右孩子为根节点的子树重复同样的过程。

这个算法满足O(h)的空间复杂度，，hasNext()的时间复杂度满足O(1)，尽管next()的时间复杂度为O(h)，但是这个算法的处理过程仍然值得学习。

C++代码实现如下：

/** * 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(); */

该算法的时间性能表现如下：

自然而然不想去因为别人的努力而努力，