【LeetCode】173. Binary Search Tree Iterator (2 solutions)

Binary Search Tree Iterator

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.

解法一：暴力解法先不考虑空间复杂度

中序遍历后装入队列，顺序输出。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class BSTIterator {
public:
    queue<int> minq;

    map<TreeNode*, bool> m;
    stack<TreeNode *> s;
    BSTIterator(TreeNode *root) {
        //inOrder traversal
        if(root != NULL)
        {
            s.push(root);
            m[root] = true;
            while(!s.empty())
            {
                TreeNode* top = s.top();
                if(top->left && m.find(top->left) == m.end())
                {
                    s.push(top->left);
                    m[top->left] = true;
                    continue;
                }
                minq.push(top->val);
                s.pop();
                if(top->right && m.find(top->right) == m.end())
                {
                    s.push(top->right);
                    m[top->right] = true;
                }
            }
        }
    }

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

    /** @return the next smallest number */
    int next() {
        int front = minq.front();
        minq.pop();
        return front;
    }
};

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

解法二：空间复杂度O(h)的解法

每次取出栈顶元素（即当前最小）后，查找下一个元素并压栈。

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

    /** @return whether we have a next smallest number */
    bool hasNext() {
        if(!stk.empty())
        {
            TreeNode* top = stk.top();
            stk.pop();
            nextmin = top->val;
            TreeNode* cur = top->right;
            if(cur)
            {
                stk.push(cur);
                cur = cur->left;
                while(cur)
                {
                    stk.push(cur);
                    cur = cur->left;
                }
            }
            return true;
        }
        else
            return false;
    }

    /** @return the next smallest number */
    int next() {
        return nextmin;
    }
};

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