173. Binary Search Tree Iterator （Medium)

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.

Solution:

这道题主要就是考二叉树的中序遍历的非递归形式，需要额外定义一个stack来辅助

use inorder search

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class BSTIterator {
public:
    // do an inorder traversal
    BSTIterator(TreeNode *root) {
        while (root) {
            s.push(root);
            root = root->left;
        }
    }

    /** @return whether we have a next smallest number */
    bool hasNext() {
        return !s.empty();
    }

    /** @return the next smallest number */
    int next() {
        TreeNode* temp = s.top(); s.pop();
        int val = temp->val;
        if (temp->right) {
           temp = temp->right;
           while (temp) {
                s.push(temp);
                temp = temp->left; 
           }
        }
        return val; 
    }
private:
    stack<TreeNode*> s;
};

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