LeetCode 297: Serialize and Deserialize Binary Tree Solution

Master LeetCode problem 297 (Serialize and Deserialize Binary Tree), a hard 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.

297. Serialize and Deserialize Binary Tree

Hard

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

Problem Explanation

Explanation

To serialize a binary tree, we can perform a preorder traversal (root-left-right) and store the node values in a string. For deserialization, we can recreate the original tree structure by parsing the serialized string. We use a delimiter to separate node values in the serialized string.

Serialization:

Serialize the root node value.
Recursively serialize the left subtree.
Recursively serialize the right subtree.

Deserialization:

Split the serialized string by the delimiter to get node values.
Reconstruct the tree using the node values in preorder traversal order.

Time Complexity: O(N) where N is the number of nodes in the tree. Space Complexity: O(N) for both serialization and deserialization.

Solution Code

public class Codec {

    private static final String DELIMITER = ",";

    // Encodes a tree to a single string.
    public String serialize(TreeNode root) {
        StringBuilder sb = new StringBuilder();
        serializeHelper(root, sb);
        return sb.toString();
    }

    private void serializeHelper(TreeNode node, StringBuilder sb) {
        if (node == null) {
            sb.append("null").append(DELIMITER);
            return;
        }

        sb.append(node.val).append(DELIMITER);
        serializeHelper(node.left, sb);
        serializeHelper(node.right, sb);
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        Queue<String> nodes = new LinkedList<>(Arrays.asList(data.split(DELIMITER)));
        return deserializeHelper(nodes);
    }

    private TreeNode deserializeHelper(Queue<String> nodes) {
        String val = nodes.poll();
        if (val.equals("null")) {
            return null;
        }

        TreeNode node = new TreeNode(Integer.parseInt(val));
        node.left = deserializeHelper(nodes);
        node.right = deserializeHelper(nodes);
        return node;
    }
}

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