[leetcode]173. Binary Search Tree Iterator

Analysis

週一鴨—— [每天刷題並不難0.0]

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.
非遞迴的方式實現二叉樹的中序遍歷~

Implement

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

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

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

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