Binary Search Tree Iterator

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.

解析：

使用栈结构，开始时存放从根节点沿着左子树经过的所有节点，当前最小的元素即是栈顶元素，下一个最小元素即是把栈顶元素的右子树节点压进栈，并把该右子树节点地所有沿着左子树的节点压进栈。

代码：

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