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Copy pathBSTIterator.go
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92 lines (75 loc) · 2.01 KB
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/* https://leetcode.com/problems/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.
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.
*/
package ldesign
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
type TreeNode struct {
Val int
Left *TreeNode
Right *TreeNode
}
type BSTIterator struct {
stack []*TreeNode
}
func BSTIteratorConstructor(root *TreeNode) BSTIterator {
bst := BSTIterator{stack: []*TreeNode{}}
bst.inOrder(root)
return bst
}
func (this *BSTIterator) inOrder(root *TreeNode) {
if root == nil {
return
}
this.stack = append(this.stack, root)
this.inOrder(root.Left)
}
/** @return the next smallest number */
func (this *BSTIterator) Next() int {
last := len(this.stack) - 1
next := this.stack[last]
this.stack = this.stack[:last]
if next.Right != nil {
this.inOrder(next.Right)
}
return next.Val
}
/** @return whether we have a next smallest number */
func (this *BSTIterator) HasNext() bool {
if len(this.stack) != 0 {
return true
}
return false
}
/**
* Your BSTIterator object will be instantiated and called as such:
* obj := Constructor(root);
* param_1 := obj.Next();
* param_2 := obj.HasNext();
*/