序列化和反序列化二叉树

小灰灰 2022-05-24 08:54 233阅读 0赞

![这里写图片描述][70]

import java.util.*;
    
     class TreeNode { 
        int val = 0;
        TreeNode left = null;
        TreeNode right = null;
    
        public TreeNode(int val) {
            this.val = val;
    
        }
    
    }
    
    public class Solution { 
          int index=-1;
        //序列化，将对象转换成字符串
       String Serialize(TreeNode root) {
            StringBuilder sb=new StringBuilder();
            if(root==null){
                sb.append("#,");
                return sb.toString();
            }
            sb.append(root.val+",");
            sb.append(Serialize(root.left));
            sb.append(Serialize(root.right));
            return sb.toString();
       }
       //反序列化
       TreeNode Deserialize(String str) {
            index++;
            String[]arr=str.split(",");
            TreeNode node=null;
            if(!arr[index].equals("#")){
    
                 node=new TreeNode(Integer.valueOf(arr[index]));
                 node.left=Deserialize(str);
                 node.right=Deserialize(str);
            }
            return node;
    
      }
       //二叉树的层序遍历
        public String TreeBFS(TreeNode pRoot,String str)
        {
         Queue<TreeNode> arr=new LinkedList<TreeNode>();//存放元素的队列
         arr.offer(pRoot);
         boolean flag=false;
         while(!arr.isEmpty())
         {
               TreeNode temp=arr.poll();
              // System.out.println(temp.val);
               str+=temp.val;
                if(temp.left!=null)
                {
                   arr.offer(temp.left);
                }
                if(temp.right!=null)
                {
                   arr.offer(temp.right);
                }
    
         }
         System.out.println(str);
          return str;
    
        }
        public static void main(String[]args){
             //System.out.println("Hello");
            TreeNode root=new TreeNode(1);
            root.left=new TreeNode(2);
            root.right=new TreeNode(3);
            root.left.left=new TreeNode(4);
            root.left.right=new TreeNode(5);
            root.right.left=new TreeNode(6);
            root.right.right=new TreeNode(7);
            Solution s=new Solution();
            String str="";
            s.TreeBFS(root,str);
    
    
        }
    
    }

[70]: /images/20220524/8946184a49dc4ba4be0715cc5b926419.png