LeetCode（173） Binary Search Tree Iterator

题目

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.

分析

实现二叉查找树的迭代器；

如题目描述，迭代器包括hasNext()和next()两个函数，其中hasNext()函数判断是否还有下一节点，next()函数返回节点元素值；且遍历顺序按照元素递增方式；

我们知道，对于二叉查找树的中序遍历结果为递增有序；所以此题的简单解法即是得到该二叉查找树的中序遍历序列并用queue保存；然后对queue进行处理；

AC代码

/**
 * 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) {   
        //中序遍历该二叉查找树，得到有序序列
        inOrder(root, inOrderNodes);
    }

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

    /** @return the next smallest number */
    int next() {
        TreeNode *node = inOrderNodes.front();
        inOrderNodes.pop();
        return node->val;
    }
private:
    queue<TreeNode *> inOrderNodes;

    void inOrder(TreeNode *root, queue<TreeNode *> &inOrderNodes)
    {
        if (!root)
            return;
        if (root->left)
            inOrder(root->left, inOrderNodes);

        inOrderNodes.push(root);

        if (root->right)
            inOrder(root->right, inOrderNodes);
    }
};

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