leetcode解题: Binary Search Tree Iterator (173)

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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.

解法1： Stack

用一个stack先预存排序了的treenode， 每次要调用next（）和hasNext（）的时候，只需要对stack操作即可。
C++

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/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class BSTIterator {
private:
    stack<TreeNode*> data;
public:
    BSTIterator(TreeNode *root) {
        traverse(root);
    }
    
    void traverse(TreeNode* root) {
        if (!root) {
            return;
        }
        
        if (!root->right && !root->left) {
            data.push(root);
            return;
        }
        
        traverse(root->right);
        data.push(root);
        traverse(root->left);
    }
    /** @return whether we have a next smallest number */
    bool hasNext() {
        return !data.empty();
    }
    /** @return the next smallest number */
    int next() {
        TreeNode* temp = data.top();
        int res = temp->val;
        data.pop();
        return res;
    }
};
/**
 * Your BSTIterator will be called like this:
 * BSTIterator i = BSTIterator(root);
 * while (i.hasNext()) cout << i.next();
 */

Java

1

解法2： Stack， 非递归的in order traversal

实际上不需要预先存储排了序的treenode，而是在寻找最小的值的时候把经过的node用stack存起来。
这里实际上考的是iterative的in order traversal

C++

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/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class BSTIterator {
private:
    stack<TreeNode*> data;
    TreeNode* cur;
public:
    BSTIterator(TreeNode *root) {
        cur = root;
    }
    /** @return whether we have a next smallest number */
    bool hasNext() {
        return cur || !data.empty();
    }
    /** @return the next smallest number */
    int next() {
        while (cur) {
            data.push(cur);
            cur = cur->left;
        }
        TreeNode* temp = data.top();
        int res = temp->val;
        data.pop();
        cur = temp->right;
        return res;
    }
};
/**
 * Your BSTIterator will be called like this:
 * BSTIterator i = BSTIterator(root);
 * while (i.hasNext()) cout << i.next();
 */