[LeetCode] 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.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

Thanks to xcv58，无需预先进行全部遍历。

利用中序遍历的递归思想：当子树的根节点访问完成后，后续节点为中序遍历该根节点右子树

/**
 * 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) {
        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;
    }
    
    void pushLeft(TreeNode *root) {
        if (root != NULL) {
            TreeNode *p = root;
            st.push(p);
            while (p->left) {
                st.push(p->left);
                p = p->left;
            }
        }
    }
};

/**
 * Your BSTIterator will be called like this:
 * BSTIterator i = BSTIterator(root);
 * while (i.hasNext()) cout << i.next();
 */