LeetCode 449: Serialize and Deserialize BST Solution

Master LeetCode problem 449 (Serialize and Deserialize BST), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

449. Serialize and Deserialize BST

Medium

String Tree Depth-First Search Breadth-First Search Design Binary Search Tree Binary Tree

Problem Explanation

Explanation

To serialize and deserialize a binary search tree (BST), we can perform an in-order traversal to serialize the BST into a string. During deserialization, we can reconstruct the BST using the in-order traversal string. We will use recursion for both serialization and deserialization.

During Serialization:
- Perform an in-order traversal of the BST.
- Append each node's value followed by a delimiter to the serialized string.
During Deserialization:
- Split the input string to extract the values of the nodes.
- Recursively build the BST using the extracted values.

Solution Code

public class Codec {

    public String serialize(TreeNode root) {
        StringBuilder sb = new StringBuilder();
        serializeHelper(root, sb);
        return sb.toString();
    }

    private void serializeHelper(TreeNode root, StringBuilder sb) {
        if (root == null) {
            return;
        }
        serializeHelper(root.left, sb);
        sb.append(root.val).append(",");
        serializeHelper(root.right, sb);
    }

    public TreeNode deserialize(String data) {
        String[] values = data.split(",");
        return deserializeHelper(values, 0, values.length - 1);
    }

    private TreeNode deserializeHelper(String[] values, int start, int end) {
        if (start > end) {
            return null;
        }
        TreeNode node = new TreeNode(Integer.parseInt(values[start]));
        int index;
        for (index = start; index <= end; index++) {
            if (Integer.parseInt(values[index]) > node.val) {
                break;
            }
        }
        node.left = deserializeHelper(values, start + 1, index - 1);
        node.right = deserializeHelper(values, index, end);
        return node;
    }
}

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