LeetCode

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.

构造一个迭代器，每次查询二叉搜索树中下一个最小值。

用一个栈来构造，每次把一个点的所有左子树丢进去。next()可以实现O(1)time，hasNext()是O(h)time。空间复杂度O(h)

/** * 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*> st;    BSTIterator(TreeNode *root) {        solve_left(root);    }    /** @return whether we have a next smallest number */    bool hasNext() {        return !st.empty();    }    /** @return the next smallest number */    int next() {        TreeNode* cur = st.top();        st.pop();        if (cur->right) {            solve_left(cur->right);        }        return cur->val;    }        void solve_left(TreeNode* root) {        while (root) {            st.push(root);            root = root->left;        }        return;    }};/** * Your BSTIterator will be called like this: * BSTIterator i = BSTIterator(root); * while (i.hasNext()) cout << i.next(); */