297. Serialize and Deserialize Binary Tree

Problem Statement

Pattern: Related: 652. Find Duplicate Subtrees

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Solution

// preorder build string fun
void preString(TreeNode root, StringBuilder sb) {
	if (root == null) {
		sb.append("#").append(" ");
		return;
	}
	sb.append(root.val).append(" ");
	preString(root.left, sb);
	preString(root.right, sb);
}
 
// preorder build nodes fun
TreeNode preNode(Queue<String> nodes) {
	String val = nodes.remove();
	if (val.equals("#")) return null;
	
	TreeNode root = new TreeNode(Integer.parseInt(val));
	root.left = preNode(nodes);
	root.right = preNode(nodes);
	
	return root;
}
 
// Encodes a tree to a single string.
public String serialize(TreeNode root) {
	StringBuilder sb = new StringBuilder();
	preString(root, sb);
	return sb.toString();
}
 
// Decodes your encoded data to tree.
public TreeNode deserialize(String str) {
	Queue<String> nodes = new LinkedList<>(Arrays.asList(str.split(" ")));
	return preNode(nodes);
}

Notes

• preorder traverse, build string
• preorder traverse, build tree 🤷