Leetcode 173. 二叉搜索树迭代器

题目连接

解题思路

根据题目当中要求返回的迭代顺序，我们可以推倒出是按后序遍历的逆序进行返回的。所以我们先根据后序遍历的方式存储二叉树的节点到数组中，然后再将此数组进行逆序，最后将结果返回。

复杂度分析

O(n)

示例代码

type TreeNode struct {
   Val   int
   Left  *TreeNode
   Right *TreeNode
}


type Stack []*TreeNode


func (stack Stack) Len() int {
   return len(stack)
}


func (stack *Stack) Push(value *TreeNode) {
   *stack = append(*stack, value)
}


func (stack Stack) Top() *TreeNode {
   if len(stack) == 0 {
      return nil
   }
   return stack[len(stack)-1]
}


func (stack *Stack) Pop() *TreeNode {
   theStack := *stack
   if len(theStack) == 0 {
      return &TreeNode{}
   }
   value := theStack[len(theStack)-1]
   *stack = theStack[:len(theStack)-1]
   return value
}


type BSTIterator struct {
   content []int
}


func Constructor(root *TreeNode) BSTIterator {
   iterator := BSTIterator{
      content: []int{},
   }
   stack := new(Stack)
   cur := root


   for stack.Len() != 0 || cur != nil {
      if cur != nil {
         stack.Push(cur)
         cur = cur.Right
      } else {
         node := stack.Pop()
         iterator.content = append(iterator.content, node.Val)
         cur = node.Left
      }
   }


   for i := 0; i < len(iterator.content)/2; i++ {
      iterator.content[i], iterator.content[len(iterator.content)-i-1] = iterator.content[len(iterator.content)-i-1], iterator.content[i]
   }


   return iterator
}


func (this *BSTIterator) Next() int {
   val := this.content[0]
   this.content = this.content[1:]
   return val
}


func (this *BSTIterator) HasNext() bool {
   return len(this.content) != 0
}