173. Binary Search Tree Iterator

題目：

173. Binary Search Tree Iterator

Medium

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Implement the BSTIterator class that represents an iterator over the in-order traversal of a binary search tree (BST):

• BSTIterator(TreeNode root) Initializes an object of the BSTIterator class. The root of the BST is given as part of the constructor. The pointer should be initialized to a non-existent number smaller than any element in the BST.
• boolean hasNext() Returns true if there exists a number in the traversal to the right of the pointer, otherwise returns false.
• int next() Moves the pointer to the right, then returns the number at the pointer.

Notice that by initializing the pointer to a non-existent smallest number, the first call to next() will return the smallest element in the BST.

You may assume that next() calls will always be valid. That is, there will be at least a next number in the in-order traversal when next() is called.

 

Example 1:

Input
["BSTIterator", "next", "next", "hasNext", "next", "hasNext", "next", "hasNext", "next", "hasNext"]
[[[7, 3, 15, null, null, 9, 20]], [], [], [], [], [], [], [], [], []]
Output
[null, 3, 7, true, 9, true, 15, true, 20, false]

Explanation
BSTIterator bSTIterator = new BSTIterator([7, 3, 15, null, null, 9, 20]);
bSTIterator.next();    // return 3
bSTIterator.next();    // return 7
bSTIterator.hasNext(); // return True
bSTIterator.next();    // return 9
bSTIterator.hasNext(); // return True
bSTIterator.next();    // return 15
bSTIterator.hasNext(); // return True
bSTIterator.next();    // return 20
bSTIterator.hasNext(); // return False

 

Constraints:

• The number of nodes in the tree is in the range [1, 10^5].
• 0 <= Node.val <= 106
• At most 105 calls will be made to hasNext, and next.

 

Follow up:

• Could you implement next() and hasNext() to run in average O(1) time and use O(h) memory, where h is the height of the tree?

 

思路：

中序遍歷，左中右。因為代替遞迴，直接想到stack，也符合follow up裡的時間和空間。思路其實就是用棧模擬中序遍歷，有左就一直往左，並且把node加入stack，每次的next就是棧頂node的值。然後依據棧頂的node，如果有右，則pop當前棧頂，把當前右子壓入棧，並且再次一直往左走即可。

 

 

 

程式碼：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class BSTIterator {
public:
    BSTIterator(TreeNode* root) {
        while(root)
        {
            s.push(root);
            root=root->left;
        }
    }
    
    int next() {
        auto t=s.top();
        int ans=t->val;
        s.pop();
        t=t->right;
        while(t)
        {
            s.push(t);
            t=t->left;
        }
        return ans;
    }
    
    bool hasNext() {
        return s.size()?true:false;
    }
private:
    stack<TreeNode*> s;
};

/**
 * Your BSTIterator object will be instantiated and called as such:
 * BSTIterator* obj = new BSTIterator(root);
 * int param_1 = obj->next();
 * bool param_2 = obj->hasNext();
 */