Skip to content
This repository was archived by the owner on Aug 14, 2024. It is now read-only.

Latest commit

 

History

History
84 lines (64 loc) · 2.2 KB

297.md

File metadata and controls

84 lines (64 loc) · 2.2 KB

297. Serialize and Deserialize Binary Tree

Description

Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.

Design an algorithm to serialize and deserialize a binary tree. There is no restriction on how your serialization/deserialization algorithm should work. You just need to ensure that a binary tree can be serialized to a string and this string can be deserialized to the original tree structure.

For example, you may serialize the following tree

    1
   / \
  2   3
     / \
    4   5

as

"[1,2,3,null,null,4,5]"

, just the same as

how LeetCode OJ serializes a binary tree

. You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.

Thinking Process

  1. Encode the tree using DFS
  2. Decode the tree by reverse DFS

Code

public class Codec {

    // Encodes a tree to a single string.
    public String serialize(TreeNode root) {
        StringBuilder sb = new StringBuilder();
        dfs(root, sb);
        return sb.toString();
    }
    
    void dfs(TreeNode root, StringBuilder sb){
        if(root == null){
            sb.append("n ");
        }else{
            sb.append(root.val + " ");
            dfs(root.left, sb);
            dfs(root.right, sb);
        }
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        String[] nodes = data.split(" ");
        int[] index = new int[1];
        return reverseDFS(nodes, index);
    }
    
    TreeNode reverseDFS(String[] nodes, int[] index){
        if(index[0] >= nodes.length){
            return null;
        }else{
            if(nodes[index[0]].equals("n")){
                index[0]++;
                return null;
            }
            else{
                TreeNode root = new TreeNode(Integer.valueOf(nodes[index[0]++]));
                root.left = reverseDFS(nodes, index);
                root.right = reverseDFS(nodes, index);
                return root;
            }
        }
    }
}