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Problem solution · Java

Serialize and Deserialize BST

Serialize and Deserialize BST: a Java solution using breadth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Breadth-first search
Source
walkccc LeetCode Solutions
Length
48 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Breadth-first search

For Serialize and Deserialize BST, the implementation explores reachable states in layers, which is the standard shape for unweighted shortest paths and minimum-step transitions.

  1. Model each valid configuration as a state and each legal move as an edge.
  2. Seed the queue with the starting state and mark it immediately.
  3. Expand each state once, recording distance or reachability for unseen neighbours.

Code notes

  • 48 lines of Java from the credited upstream file 449.java.
  • The implementation visibly relies on sequence storage, work queue.
  • 1 loop block detected.

Complexity

Verify that each state and transition is processed only a bounded number of times; that determines the traversal cost.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeSerialize and Deserialize BST · JavaJava
Use this to learn the idea, then write your own version.
public class Codec {  // Encodes a tree to a single string.  public String serialize(TreeNode root) {    if (root == null)      return "";    StringBuilder sb = new StringBuilder();    serialize(root, sb);    return sb.toString();  }   // Decodes your encoded data to tree.  public TreeNode deserialize(String data) {    if (data.isEmpty())      return null;     final String[] vals = data.split(" ");    Queue<Integer> q = new ArrayDeque<>();     for (final String val : vals)      q.offer(Integer.parseInt(val));     return deserialize(Integer.MIN_VALUE, Integer.MAX_VALUE, q);  }   private void serialize(TreeNode root, StringBuilder sb) {    if (root == null)      return;    sb.append(root.val).append(" ");    serialize(root.left, sb);    serialize(root.right, sb);  }   private TreeNode deserialize(int mn, int mx, Queue<Integer> q) {    if (q.isEmpty())      return null;     final int val = q.peek();    if (val < mn || val > mx)      return null;     q.poll();    TreeNode root = new TreeNode(val);    root.left = deserialize(mn, val, q);    root.right = deserialize(val, mx, q);    return root;  }}

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