LeetCode #173 - Binary Search Tree Iterator - Medium

Problem

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.

Example

Given a binary search tree below and its root    4   / \  2   5 / \1   3 with > BSTIterator i = BSTIterator(root); > while (i.hasNext()) cout << i.next();output: 1, 2, 3, 4, 5

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Algorithm

整理一下题意：给定一棵二叉搜索树，定义函数next()返回二叉搜索树中下一个最小值，hasnext()函数返回这棵二叉树是否有下一个最小值。

由于二叉搜索树的特点，左子<父<右子，所以利用栈中序遍历整棵树即可。题中拆分的几个函数实际上是将中序遍历的过程拆分成三个阶段（有一个函数需要自己补全）。

代码如下。

class BSTIterator {public:    BSTIterator(TreeNode *root) {        pushLeft(root);    }    /** @return whether we have a next smallest number */    bool hasNext() {        return !st.empty();    }    /** @return the next smallest number */    int next() {        TreeNode*top=st.top();        st.pop();        pushLeft(top->right);        return top->val;    }private:    stack<TreeNode*> st;    void pushLeft(TreeNode* root){        while(root!=NULL){            st.push(root);            root=root->left;        }    }};