Leetcode-173 Binary-Search-Tree-Iterator

Description

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.

Example:

BSTIterator iterator = new BSTIterator(root);
iterator.next();    // return 3
iterator.next();    // return 7
iterator.hasNext(); // return true
iterator.next();    // return 9
iterator.hasNext(); // return true
iterator.next();    // return 15
iterator.hasNext(); // return true
iterator.next();    // return 20
iterator.hasNext(); // return false
 

Note:

• next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.
• You may assume that next() call will always be valid, that is, there will be at least a next smallest number in the BST when next() is called.

Answer:

第一反应还是用栈来做。先递归把左子树全部压进去，之后每调用一次next()，就出栈，同时把出栈的所有左路径上的节点压入。
而hasNext()则是返回栈是否为空。

下面是代码：

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

    BSTIterator(TreeNode* root) {
    
    
        push(root);
    }
    
    /** @return the next smallest number */
    int next() {
    
    
        TreeNode* temp = sta.top();
        sta.pop();
        if (temp -> right)
            push(temp -> right);
        return temp -> val;
    }
    
    /** @return whether we have a next smallest number */
    bool hasNext() {
    
    
        return !sta.empty();
    }
private:
    stack<TreeNode*> sta;
    int size;
    int pos;
};

/**
 * 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();
 */

上述next()的空间复杂度是O(h)，而通过摊还分析，我们也可以计算出其时间复杂度是O(1)，所以是符合题目要求的。

不过之后想了一想，既然题目至少要求next()和hasNext()的时间复杂度和空间复杂度，那么构造函数的复杂度就是没有限制的。所以我们可以直接在构造函数中进行中序遍历，这样就可以保证next()和hasNext()的空间和时间都是常数复杂度了。

下面是代码：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class BSTIterator {
    
    
public:
    vector<int> arr;
    int pos;
    
    void compute(TreeNode* root) {
    
    
        if (!root) 
            return;
        compute(root -> left);
        arr.push_back(root -> val);
        compute(root -> right);
        
    }
    BSTIterator(TreeNode* root) {
    
    
        compute(root);
        pos = 0;
    }
    
    /** @return the next smallest number */
    int next() {
    
    
        return arr[pos++];
    }
    
    /** @return whether we have a next smallest number */
    bool hasNext() {
    
    
        return pos < arr.size();
    }
};

/**
 * 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();
 */