173. Binary Search Tree Iterator

Total Accepted: 50775  Total Submissions: 144066  Difficulty: 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.

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

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分析：

感觉蛮简单的！就是一个中序遍历嘛！可能想的太简单了！

next()和hasnext()函数的空间复杂度为O(1)，时间复杂度为O(1)，这里并没有符合要求。

/**
 * 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) {
        inorderTraversal(root);//获取中序遍历结果
        pos=0;
    }
    void inorderTraversal(TreeNode* root) { 
         if(root){      
            inorderTraversal(root->left);      
            result.push_back(root->val);      
            inorderTraversal(root->right);      
        }      
    }      
    /** @return whether we have a next smallest number */
    bool hasNext() {
        return pos < result.size();
    }

    /** @return the next smallest number */
    int next() {
        return result[pos++];
    }
private:      
    vector<int> result;
    int pos;
};

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

别人的算法设计：

class BSTIterator {
    
public:
    BSTIterator(TreeNode *root) {
        pushLeft(root);
    }

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

    /** @return the next smallest number */
    int next() {//空间复杂度为O(h)，时间复杂度为O(h)
        TreeNode *tmpNode = stk.top();
        stk.pop();
        pushLeft(tmpNode->right);
        return tmpNode->val;
    }
    
    void pushLeft(TreeNode *node) {//空间复杂度为O(h)，时间复杂度为O(h)
        while (node != NULL)
        {
            stk.push(node);
            node = node->left;
        }
    }
    
private:
    stack<TreeNode *> stk;
};

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原文地址：http://blog.csdn.net/ebowtang/article/details/51660101

原作者博客：http://blog.csdn.net/ebowtang

本博客LeetCode题解索引：http://blog.csdn.net/ebowtang/article/details/50668895