Binary Search Tree Iterato

一、问题描述

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.

二、思路

在开始定义一个头指针，定义一个类的构造函数，hasnext（）函数很简单，判断头指针是否为空，为空返回false，不为空返回true。

在next（）函数中找到最左侧的元素即为BST树最小的元素，当pre不为空时，使pre的左节点指向当前节点的右子树；当pre为空时，pre的左节点指向head。

最后返回最小值。

三、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:    TreeNode *head;    BSTIterator(TreeNode *root) {        head=root;    }    /** @return whether we have a next smallest number */    bool hasNext() {        if(!head)            return false;        return true;    }    /** @return the next smallest number */    int next() {        TreeNode *cur = head;        TreeNode *pre = NULL;        while(cur -> left){            pre = cur;            cur = cur -> left;        }        int temp = cur -> val;        if(pre)            pre -> left = cur -> right;        else            head = cur -> right;        return temp;    }};/** * Your BSTIterator will be called like this: * BSTIterator i = BSTIterator(root); * while (i.hasNext()) cout << i.next(); */