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118 lines
2.6 KiB
Markdown
Executable File
118 lines
2.6 KiB
Markdown
Executable File
# [173. Binary Search Tree Iterator](https://leetcode.com/problems/binary-search-tree-iterator/)
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## 题目
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Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST.
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Calling `next()` will return the next smallest number in the BST.
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**Example**:
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BSTIterator iterator = new BSTIterator(root);
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iterator.next(); // return 3
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iterator.next(); // return 7
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iterator.hasNext(); // return true
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iterator.next(); // return 9
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iterator.hasNext(); // return true
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iterator.next(); // return 15
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iterator.hasNext(); // return true
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iterator.next(); // return 20
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iterator.hasNext(); // return false
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**Note**:
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- `next()` and `hasNext()` should run in average O(1) time and uses O(h) memory, where h is the height of the tree.
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- 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.
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## 题目大意
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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。
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## 解题思路
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- 用优先队列解决即可
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## 代码
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```go
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package leetcode
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import "container/heap"
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/**
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* Definition for a binary tree node.
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* type TreeNode struct {
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* Val int
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* Left *TreeNode
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* Right *TreeNode
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* }
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*/
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// BSTIterator define
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type BSTIterator struct {
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pq PriorityQueueOfInt
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count int
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}
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// Constructor173 define
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func Constructor173(root *TreeNode) BSTIterator {
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result, pq := []int{}, PriorityQueueOfInt{}
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postorder(root, &result)
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for _, v := range result {
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heap.Push(&pq, v)
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}
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bs := BSTIterator{pq: pq, count: len(result)}
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return bs
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}
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/** @return the next smallest number */
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func (this *BSTIterator) Next() int {
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this.count--
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return heap.Pop(&this.pq).(int)
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}
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/** @return whether we have a next smallest number */
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func (this *BSTIterator) HasNext() bool {
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return this.count != 0
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}
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/**
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* Your BSTIterator object will be instantiated and called as such:
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* obj := Constructor(root);
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* param_1 := obj.Next();
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* param_2 := obj.HasNext();
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*/
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type PriorityQueueOfInt []int
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func (pq PriorityQueueOfInt) Len() int {
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return len(pq)
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}
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func (pq PriorityQueueOfInt) Less(i, j int) bool {
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return pq[i] < pq[j]
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}
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func (pq PriorityQueueOfInt) Swap(i, j int) {
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pq[i], pq[j] = pq[j], pq[i]
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}
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func (pq *PriorityQueueOfInt) Push(x interface{}) {
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item := x.(int)
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*pq = append(*pq, item)
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}
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func (pq *PriorityQueueOfInt) Pop() interface{} {
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n := len(*pq)
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item := (*pq)[n-1]
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*pq = (*pq)[:n-1]
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return item
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}
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``` |