[LeetCode] Binary Search Tree Iterator

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.

 

解题思路：

运用栈的技术，将当前未被访问的最左边的一条路径入栈，每次取值的时候将栈顶元素弹出，并将栈顶元素的右子树的最左边一条路径入栈。hasNext()只需要看栈是否为空即可。但是有个问题，似乎next()函数的时间复杂度并不是O(1)，因为我们要维护栈，因此是O(nlgn)。不知有什么更为好的方法没有。我还犯了一个小错误，就是stack<TreeNode*> stack。不能取名叫stack呀。

/**
 * 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) {
        pushLeftChildIntoStack(root);
    }

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

    /** @return the next smallest number */
    int next() {
        TreeNode* node=s.top();
        s.pop();
        pushLeftChildIntoStack(node->right);
        return node->val;
    }
private:
    stack<TreeNode*> s;
    void pushLeftChildIntoStack(TreeNode* node){
        while(node!=NULL){
            s.push(node);
            node=node->left;
        }
    }
};

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

[LeetCode] Binary Search Tree Iterator

原文：http://blog.csdn.net/kangrydotnet/article/details/44986675