Approach Summary
Pre-order DFS serialization with null markers. Deserialization reconstructs using a queue of values.
Practice Mode
3 Progressive Approaches
Solve it yourself before revealing the code
DSA interviews test reasoning from brute force to optimal. Think through the constraints first.
Solution Progression (Brute Force → Most Optimized)
Understand the architectural evolution from naive baseline to optimal complexity.
| Approach | Time | Space | Core Strategy / Trade-off |
|---|---|---|---|
| 1 Approach 1: BFS Level-Order with Delimiters (Optimal) | O(n) | O(n) | Encode level-by-level with null markers (#); reconstruct children pairs using a FIFO queue. |
| 2 Approach 2: Breadth-First Search (BFS) Level-Order with Queue | O(n) | O(n) | Serializes using BFS queue inserting "null" tokens, deserializing with parent queue. |
| 3 Approach 2: Breadth-First Search (BFS) Level-Order with Queue | O(n) | O(n) | Serializes using BFS queue inserting "null" tokens, deserializing with parent queue. |
Step 1: Baseline
Approach 1: BFS Level-Order with Delimiters (Optimal)
Time: O(n)
Space: O(n)
Level-order traversal naturally captures tree structure when null leaf pointers are recorded as '#'. Upon decoding, reading tokens in sequential pairs assigns the left and right children of each dequeued parent node.
Approach 1: BFS Level-Order with Delimiters (Optimal) Implementation
Approach 1: BFS Level-Order with Delimiters (Optimal)
public class Codec {
public String serialize(TreeNode root) {
if (root == null) return "";
StringBuilder sb = new StringBuilder();
Queue<TreeNode> q = new LinkedList<>();
q.add(root);
while (!q.isEmpty()) {
TreeNode n = q.poll();
if (n == null) sb.append("#,");
else { sb.append(n.val).append(","); q.add(n.left); q.add(n.right); }
}
return sb.toString();
}
public TreeNode deserialize(String data) {
if (data.isEmpty()) return null;
String[] vals = data.split(",");
TreeNode root = new TreeNode(Integer.parseInt(vals[0]));
Queue<TreeNode> q = new LinkedList<>();
q.add(root);
for (int i = 1; i < vals.length; i++) {
TreeNode p = q.poll();
if (!vals[i].equals("#")) { TreeNode l = new TreeNode(Integer.parseInt(vals[i])); p.left = l; q.add(l); }
if (++i < vals.length && !vals[i].equals("#")) { TreeNode r = new TreeNode(Integer.parseInt(vals[i])); p.right = r; q.add(r); }
}
return root;
}
}class Codec {
public:
string serialize(TreeNode* root) {
if (!root) return "";
string s = ""; queue<TreeNode*> q; q.push(root);
while (!q.empty()) {
TreeNode* n = q.front(); q.pop();
if (!n) s += "#,";
else { s += to_string(n->val) + ","; q.push(n->left); q.push(n->right); }
}
return s;
}
TreeNode* deserialize(string data) {
if (data.empty()) return NULL;
vector<string> v; string cur = "";
for (char c : data) { if (c == ',') { v.push_back(cur); cur = ""; } else cur += c; }
TreeNode* root = new TreeNode(stoi(v[0]));
queue<TreeNode*> q; q.push(root);
for (int i = 1; i < v.size(); i++) {
TreeNode* p = q.front(); q.pop();
if (v[i] != "#") { TreeNode* l = new TreeNode(stoi(v[i])); p->left = l; q.push(l); }
if (++i < v.size() && v[i] != "#") { TreeNode* r = new TreeNode(stoi(v[i])); p->right = r; q.push(r); }
}
return root;
}
};class Codec:
def serialize(self, root: Optional[TreeNode]) -> str:
if not root: return ""
q, res = deque([root]), []
while q:
n = q.popleft()
if n:
res.append(str(n.val))
q.append(n.left)
q.append(n.right)
else:
res.append("#")
return ",".join(res)
def deserialize(self, data: str) -> Optional[TreeNode]:
if not data: return None
vals = data.split(",")
root = TreeNode(int(vals[0]))
q, i = deque([root]), 1
while q and i < len(vals):
p = q.popleft()
if vals[i] != "#":
p.left = TreeNode(int(vals[i]))
q.append(p.left)
i += 1
if i < len(vals) and vals[i] != "#":
p.right = TreeNode(int(vals[i]))
q.append(p.right)
i += 1
return rootfunction serialize(root) {
if (!root) return "";
const q = [root], res = [];
while (q.length) {
const n = q.shift();
if (n) { res.push(n.val); q.push(n.left); q.push(n.right); }
else { res.push("#"); }
}
return res.join(",");
}
function deserialize(data) {
if (!data) return null;
const vals = data.split(","), root = new TreeNode(+vals[0]), q = [root];
let i = 1;
while (q.length && i < vals.length) {
const p = q.shift();
if (vals[i] !== "#") { p.left = new TreeNode(+vals[i]); q.push(p.left); }
i++;
if (i < vals.length && vals[i] !== "#") { p.right = new TreeNode(+vals[i]); q.push(p.right); }
i++;
}
return root;
} Step 2
Approach 2: Breadth-First Search (BFS) Level-Order with Queue
Time: O(n)
Space: O(n)
In BFS serialization, nodes are enqueued level-by-level. If a node is null, we write "null". When deserializing, split tokens and use a queue to reconstruct tree by assigning pairs of left and right children to each parent in queue.
Approach 2: Breadth-First Search (BFS) Level-Order with Queue Implementation
Approach 2: Breadth-First Search (BFS) Level-Order with Queue
public class Codec {
public String serialize(TreeNode root) {
if (root == null) return "";
Queue<TreeNode> q = new LinkedList<>();
StringBuilder sb = new StringBuilder();
q.add(root);
while (!q.isEmpty()) {
TreeNode node = q.poll();
if (node == null) {
sb.append("null,");
} else {
sb.append(node.val).append(",");
q.add(node.left);
q.add(node.right);
}
}
return sb.toString();
}
public TreeNode deserialize(String data) {
if (data == null || data.isEmpty()) return null;
String[] tokens = data.split(",");
TreeNode root = new TreeNode(Integer.parseInt(tokens[0]));
Queue<TreeNode> q = new LinkedList<>();
q.add(root);
int i = 1;
while (!q.isEmpty() && i < tokens.length) {
TreeNode parent = q.poll();
if (!tokens[i].equals("null")) {
parent.left = new TreeNode(Integer.parseInt(tokens[i]));
q.add(parent.left);
}
i++;
if (i < tokens.length && !tokens[i].equals("null")) {
parent.right = new TreeNode(Integer.parseInt(tokens[i]));
q.add(parent.right);
}
i++;
}
return root;
}
}class Codec {
public:
string serialize(TreeNode* root) {
if (!root) return "";
string res = "";
queue<TreeNode*> q;
q.push(root);
while (!q.empty()) {
TreeNode* curr = q.front(); q.pop();
if (!curr) res += "null,";
else {
res += to_string(curr->val) + ",";
q.push(curr->left);
q.push(curr->right);
}
}
return res;
}
TreeNode* deserialize(string data) {
if (data.empty()) return NULL;
stringstream ss(data);
string item;
getline(ss, item, ',');
TreeNode* root = new TreeNode(stoi(item));
queue<TreeNode*> q;
q.push(root);
while (!q.empty() && getline(ss, item, ',')) {
TreeNode* parent = q.front(); q.pop();
if (item != "null") {
parent->left = new TreeNode(stoi(item));
q.push(parent->left);
}
if (getline(ss, item, ',')) {
if (item != "null") {
parent->right = new TreeNode(stoi(item));
q.push(parent->right);
}
}
}
return root;
}
};class Codec:
def serialize(self, root):
if not root: return ""
q = collections.deque([root])
res = []
while q:
node = q.popleft()
if node:
res.append(str(node.val))
q.append(node.left)
q.append(node.right)
else:
res.append("null")
return ",".join(res)
def deserialize(self, data):
if not data: return None
tokens = collections.deque(data.split(","))
root = TreeNode(int(tokens.popleft()))
q = collections.deque([root])
while q and tokens:
parent = q.popleft()
left_val = tokens.popleft()
if left_val != "null":
parent.left = TreeNode(int(left_val))
q.append(parent.left)
right_val = tokens.popleft()
if right_val != "null":
parent.right = TreeNode(int(right_val))
q.append(parent.right)
return rootvar serialize = function(root) {
if (!root) return "";
const q = [root], res = [];
while (q.length) {
const node = q.shift();
if (node) {
res.push(node.val);
q.push(node.left);
q.push(node.right);
} else {
res.push("null");
}
}
return res.join(",");
};
var deserialize = function(data) {
if (!data) return null;
const tokens = data.split(",");
const root = new TreeNode(parseInt(tokens[0]));
const q = [root];
let i = 1;
while (q.length && i < tokens.length) {
const parent = q.shift();
if (tokens[i] !== "null") {
parent.left = new TreeNode(parseInt(tokens[i]));
q.push(parent.left);
}
i++;
if (i < tokens.length && tokens[i] !== "null") {
parent.right = new TreeNode(parseInt(tokens[i]));
q.push(parent.right);
}
i++;
}
return root;
}; Step 2: Most Optimized
Approach 2: Breadth-First Search (BFS) Level-Order with Queue
Time: O(n)
Space: O(n)
In BFS serialization, nodes are enqueued level-by-level. If a node is null, we write "null". When deserializing, split tokens and use a queue to reconstruct tree by assigning pairs of left and right children to each parent in queue.
Approach 2: Breadth-First Search (BFS) Level-Order with Queue Implementation
Approach 2: Breadth-First Search (BFS) Level-Order with Queue
public class Codec {
public String serialize(TreeNode root) {
if (root == null) return "";
Queue<TreeNode> q = new LinkedList<>();
StringBuilder sb = new StringBuilder();
q.add(root);
while (!q.isEmpty()) {
TreeNode node = q.poll();
if (node == null) {
sb.append("null,");
} else {
sb.append(node.val).append(",");
q.add(node.left);
q.add(node.right);
}
}
return sb.toString();
}
public TreeNode deserialize(String data) {
if (data == null || data.isEmpty()) return null;
String[] tokens = data.split(",");
TreeNode root = new TreeNode(Integer.parseInt(tokens[0]));
Queue<TreeNode> q = new LinkedList<>();
q.add(root);
int i = 1;
while (!q.isEmpty() && i < tokens.length) {
TreeNode parent = q.poll();
if (!tokens[i].equals("null")) {
parent.left = new TreeNode(Integer.parseInt(tokens[i]));
q.add(parent.left);
}
i++;
if (i < tokens.length && !tokens[i].equals("null")) {
parent.right = new TreeNode(Integer.parseInt(tokens[i]));
q.add(parent.right);
}
i++;
}
return root;
}
}class Codec {
public:
string serialize(TreeNode* root) {
if (!root) return "";
string res = "";
queue<TreeNode*> q;
q.push(root);
while (!q.empty()) {
TreeNode* curr = q.front(); q.pop();
if (!curr) res += "null,";
else {
res += to_string(curr->val) + ",";
q.push(curr->left);
q.push(curr->right);
}
}
return res;
}
TreeNode* deserialize(string data) {
if (data.empty()) return NULL;
stringstream ss(data);
string item;
getline(ss, item, ',');
TreeNode* root = new TreeNode(stoi(item));
queue<TreeNode*> q;
q.push(root);
while (!q.empty() && getline(ss, item, ',')) {
TreeNode* parent = q.front(); q.pop();
if (item != "null") {
parent->left = new TreeNode(stoi(item));
q.push(parent->left);
}
if (getline(ss, item, ',')) {
if (item != "null") {
parent->right = new TreeNode(stoi(item));
q.push(parent->right);
}
}
}
return root;
}
};class Codec:
def serialize(self, root):
if not root: return ""
q = collections.deque([root])
res = []
while q:
node = q.popleft()
if node:
res.append(str(node.val))
q.append(node.left)
q.append(node.right)
else:
res.append("null")
return ",".join(res)
def deserialize(self, data):
if not data: return None
tokens = collections.deque(data.split(","))
root = TreeNode(int(tokens.popleft()))
q = collections.deque([root])
while q and tokens:
parent = q.popleft()
left_val = tokens.popleft()
if left_val != "null":
parent.left = TreeNode(int(left_val))
q.append(parent.left)
right_val = tokens.popleft()
if right_val != "null":
parent.right = TreeNode(int(right_val))
q.append(parent.right)
return rootvar serialize = function(root) {
if (!root) return "";
const q = [root], res = [];
while (q.length) {
const node = q.shift();
if (node) {
res.push(node.val);
q.push(node.left);
q.push(node.right);
} else {
res.push("null");
}
}
return res.join(",");
};
var deserialize = function(data) {
if (!data) return null;
const tokens = data.split(",");
const root = new TreeNode(parseInt(tokens[0]));
const q = [root];
let i = 1;
while (q.length && i < tokens.length) {
const parent = q.shift();
if (tokens[i] !== "null") {
parent.left = new TreeNode(parseInt(tokens[i]));
q.push(parent.left);
}
i++;
if (i < tokens.length && tokens[i] !== "null") {
parent.right = new TreeNode(parseInt(tokens[i]));
q.push(parent.right);
}
i++;
}
return root;
};Edge Cases to Watch
- Empty tree (returns empty string "" or "#"; deserializes to null)
- Single node tree (e.g. "1,#,#,")
- Unbalanced or zigzag trees
Algorithmic Pattern
View Pattern Guide →
Master the Trees / DFS Pattern
Study the core invariant, step-by-step decision framework, and common interview traps for the Trees / DFS pattern.
Framework: Trees / DFS • Interview Prep: Tested at FAANG •
Explore all Trees / DFS problems & invariants →
How to Recognize This Pattern
- Encode/decode tree structure
- Pre-order with nulls
Complexity Analysis
Time Complexity
O(n)
Space Complexity
O(n)
Tags
Tree DFS BFS Design
Frequently Asked Questions
What is the optimal time and space complexity for Serialize and Deserialize Binary Tree?
The optimal solution for Serialize and Deserialize Binary Tree achieves O(n) runtime complexity and requires O(n) auxiliary space. Time complexity is O(n) for both serialization and deserialization because each node and null pointer is processed once. Space complexity is O(n) to store the serialized string and queue nodes.
How do I identify that Serialize and Deserialize Binary Tree requires the Trees / DFS pattern?
In technical interviews, look for key problem signals: Encode/decode tree structure; Pre-order with nulls. Recognizing these invariants allows you to bypass naive brute-force exhaustive search in favor of Trees / DFS strategy.
What critical edge cases must be handled in Serialize and Deserialize Binary Tree?
Essential edge cases to verify include: Empty tree (returns empty string "" or "#"; deserializes to null); Single node tree (e.g. "1,#,#,"); Unbalanced or zigzag trees. Always walk through these scenarios with your interviewer before coding.
Which tech companies frequently ask Serialize and Deserialize Binary Tree?
Serialize and Deserialize Binary Tree is a Hard-difficulty question tagged with medium recurrence. It is actively asked in technical screening and onsite interview loops at Google, Amazon, Facebook, LinkedIn.