# [173. Binary Search Tree Iterator](https://leetcode.com/problems/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. **Example**: ![](https://assets.leetcode.com/uploads/2018/12/25/bst-tree.png) BSTIterator iterator = new BSTIterator(root); iterator.next(); // return 3 iterator.next(); // return 7 iterator.hasNext(); // return true iterator.next(); // return 9 iterator.hasNext(); // return true iterator.next(); // return 15 iterator.hasNext(); // return true iterator.next(); // return 20 iterator.hasNext(); // return false **Note**: - `next()` and `hasNext()` should run in average O(1) time and uses O(h) memory, where h is the height of the tree. - You may assume that `next()` call will always be valid, that is, there will be at least a next smallest number in the BST when `next()` is called. ## 题目大意 实现一个二叉搜索树迭代器。你将使用二叉搜索树的根节点初始化迭代器。调用 next() 将返回二叉搜索树中的下一个最小的数。 ## 解题思路 - 用优先队列解决即可 ## 代码 ```go package leetcode import "container/heap" /** * Definition for a binary tree node. * type TreeNode struct { * Val int * Left *TreeNode * Right *TreeNode * } */ // BSTIterator define type BSTIterator struct { pq PriorityQueueOfInt count int } // Constructor173 define func Constructor173(root *TreeNode) BSTIterator { result, pq := []int{}, PriorityQueueOfInt{} postorder(root, &result) for _, v := range result { heap.Push(&pq, v) } bs := BSTIterator{pq: pq, count: len(result)} return bs } /** @return the next smallest number */ func (this *BSTIterator) Next() int { this.count-- return heap.Pop(&this.pq).(int) } /** @return whether we have a next smallest number */ func (this *BSTIterator) HasNext() bool { return this.count != 0 } /** * Your BSTIterator object will be instantiated and called as such: * obj := Constructor(root); * param_1 := obj.Next(); * param_2 := obj.HasNext(); */ type PriorityQueueOfInt []int func (pq PriorityQueueOfInt) Len() int { return len(pq) } func (pq PriorityQueueOfInt) Less(i, j int) bool { return pq[i] < pq[j] } func (pq PriorityQueueOfInt) Swap(i, j int) { pq[i], pq[j] = pq[j], pq[i] } func (pq *PriorityQueueOfInt) Push(x interface{}) { item := x.(int) *pq = append(*pq, item) } func (pq *PriorityQueueOfInt) Pop() interface{} { n := len(*pq) item := (*pq)[n-1] *pq = (*pq)[:n-1] return item } ```