297.二叉树的序列化与反序列化

leetcode

时间复杂度： O(n)
空间复杂度： O(n)

/**  
 * Definition for a binary tree node. * type TreeNode struct { *     Val int *     Left *TreeNode *     Right *TreeNode * } */  
type Codec struct {  
   StrArray []string  
   path int  
}  
  
func Constructor() Codec {  
   return Codec{  
      StrArray: []string{},  
      path: 0,  
   }  
}  
  
// Serializes a tree to a single string.  
func (this *Codec) serialize(root *TreeNode) string {  
   this.SeRecusival(root)  
   // fmt.Printf("array: %v\n", this.StrArray)  
  
   return strings.Join(this.StrArray, ",")  
}  
  
// Deserializes your encoded data to tree.  
func (this *Codec) deserialize(data string) *TreeNode {  
   this.StrArray = strings.Split(data, ",")  
   this.path = 0  
  
   return this.DeRecusival()  
}  
  
// 1 2 nil nil 3 4 nil nil 5 nil nil  
  
func (this *Codec) DeRecusival() *TreeNode {  
   if this.StrArray[this.path] == "nil" {  
      return nil  
   }  
  
   intVar, _ := strconv.Atoi(this.StrArray[this.path])  
  
   root := &TreeNode{Val: intVar}  
   this.path++  
   root.Left = this.DeRecusival()  
   this.path++  
   root.Right = this.DeRecusival()  
  
   return root  
}  
  
func (this *Codec) SeRecusival(root *TreeNode) {  
   if root == nil {  
      this.StrArray = append(this.StrArray, "nil")  
      return  
   }  
  
   // fmt.Printf("val: %v\n", strconv.Itoa(root.Val))  
  
   this.StrArray = append(this.StrArray, strconv.Itoa(root.Val))  
   this.SeRecusival(root.Left)  
   this.SeRecusival(root.Right)  
}  
  
  
/**

* Your Codec object will be instantiated and called as such:

* ser := Constructor();

* deser := Constructor();

* data := ser.serialize(root);

* ans := deser.deserialize(data);

*/<!-- more -->