LeetCode #449 — MEDIUM

Serialize and Deserialize BST

Move from brute-force thinking to an efficient approach using tree strategy.

The Problem

Problem Statement

Serialization is converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.

Design an algorithm to serialize and deserialize a binary search tree. There is no restriction on how your serialization/deserialization algorithm should work. You need to ensure that a binary search tree can be serialized to a string, and this string can be deserialized to the original tree structure.

The encoded string should be as compact as possible.

Example 1:

Input: root = [2,1,3]
Output: [2,1,3]

Example 2:

Input: root = []
Output: []

Constraints:

The number of nodes in the tree is in the range [0, 10⁴].
0 <= Node.val <= 10⁴
The input tree is guaranteed to be a binary search tree.

Step 01

Brute Force Baseline

Problem summary: Serialization is converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment. Design an algorithm to serialize and deserialize a binary search tree. There is no restriction on how your serialization/deserialization algorithm should work. You need to ensure that a binary search tree can be serialized to a string, and this string can be deserialized to the original tree structure. The encoded string should be as compact as possible.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Tree · Design

Example 1

[2,1,3]

Example 2

[]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #449: Serialize and Deserialize BST
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Codec {
    private int i;
    private List<String> nums;
    private final int inf = 1 << 30;

    // Encodes a tree to a single string.
    public String serialize(TreeNode root) {
        nums = new ArrayList<>();
        dfs(root);
        return String.join(" ", nums);
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        if (data == null || "".equals(data)) {
            return null;
        }
        i = 0;
        nums = Arrays.asList(data.split(" "));
        return dfs(-inf, inf);
    }

    private void dfs(TreeNode root) {
        if (root == null) {
            return;
        }
        nums.add(String.valueOf(root.val));
        dfs(root.left);
        dfs(root.right);
    }

    private TreeNode dfs(int mi, int mx) {
        if (i == nums.size()) {
            return null;
        }
        int x = Integer.parseInt(nums.get(i));
        if (x < mi || x > mx) {
            return null;
        }
        TreeNode root = new TreeNode(x);
        ++i;
        root.left = dfs(mi, x);
        root.right = dfs(x, mx);
        return root;
    }
}

// Your Codec object will be instantiated and called as such:
// Codec ser = new Codec();
// Codec deser = new Codec();
// String tree = ser.serialize(root);
// TreeNode ans = deser.deserialize(tree);
// return ans;

// Accepted solution for LeetCode #449: Serialize and Deserialize BST
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */

type Codec struct {
}

func Constructor() Codec {
	return Codec{}
}

// Serializes a tree to a single string.
func (this *Codec) serialize(root *TreeNode) string {
	if root == nil {
		return ""
	}
	data := &strings.Builder{}
	var dfs func(*TreeNode)
	dfs = func(root *TreeNode) {
		if root == nil {
			return
		}
		data.WriteString(strconv.Itoa(root.Val))
		data.WriteByte(' ')
		dfs(root.Left)
		dfs(root.Right)
	}
	dfs(root)
	return data.String()[0 : data.Len()-1]
}

// Deserializes your encoded data to tree.
func (this *Codec) deserialize(data string) *TreeNode {
	if data == "" {
		return nil
	}
	vals := strings.Split(data, " ")
	i := 0
	var dfs func(int, int) *TreeNode
	dfs = func(mi, mx int) *TreeNode {
		if i == len(vals) {
			return nil
		}
		x, _ := strconv.Atoi(vals[i])
		if x < mi || x > mx {
			return nil
		}
		i++
		root := &TreeNode{Val: x}
		root.Left = dfs(mi, x)
		root.Right = dfs(x, mx)
		return root
	}
	return dfs(math.MinInt64, math.MaxInt64)
}

/**
 * Your Codec object will be instantiated and called as such:
 * ser := Constructor()
 * deser := Constructor()
 * tree := ser.serialize(root)
 * ans := deser.deserialize(tree)
 * return ans
 */

# Accepted solution for LeetCode #449: Serialize and Deserialize BST
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None


class Codec:
    def serialize(self, root: Optional[TreeNode]) -> str:
        """Encodes a tree to a single string."""

        def dfs(root: Optional[TreeNode]):
            if root is None:
                return
            nums.append(root.val)
            dfs(root.left)
            dfs(root.right)

        nums = []
        dfs(root)
        return " ".join(map(str, nums))

    def deserialize(self, data: str) -> Optional[TreeNode]:
        """Decodes your encoded data to tree."""

        def dfs(mi: int, mx: int) -> Optional[TreeNode]:
            nonlocal i
            if i == len(nums) or not mi <= nums[i] <= mx:
                return None
            x = nums[i]
            root = TreeNode(x)
            i += 1
            root.left = dfs(mi, x)
            root.right = dfs(x, mx)
            return root

        nums = list(map(int, data.split()))
        i = 0
        return dfs(-inf, inf)


# Your Codec object will be instantiated and called as such:
# Your Codec object will be instantiated and called as such:
# ser = Codec()
# deser = Codec()
# tree = ser.serialize(root)
# ans = deser.deserialize(tree)
# return ans

// Accepted solution for LeetCode #449: Serialize and Deserialize BST
use rustgym_util::*;
use std::iter::Peekable;
use std::vec::IntoIter;

struct Codec;

enum Tok {
    Op(char),
    Num(i32),
}

impl Codec {
    fn new() -> Self {
        Codec {}
    }

    fn serialize(&self, root: TreeLink) -> String {
        let mut res = "".to_string();
        Self::serialize_tree(&root, &mut res);
        res
    }

    fn serialize_tree(root: &TreeLink, s: &mut String) {
        s.push('(');
        if let Some(node) = root {
            let node = node.borrow();
            *s += &format!("{}", node.val);
            Self::serialize_tree(&node.left, s);
            Self::serialize_tree(&node.right, s);
        }
        s.push(')');
    }

    fn deserialize(&self, data: String) -> TreeLink {
        let tokens = Self::parse_tokens(data);
        let mut it = tokens.into_iter().peekable();
        Self::parse_tree(&mut it)
    }

    fn parse_tokens(data: String) -> Vec<Tok> {
        let mut it = data.chars().peekable();
        let mut res = vec![];
        while let Some(c) = it.next() {
            if c == '(' || c == ')' {
                res.push(Tok::Op(c));
            } else {
                let mut x = (c as u8 - b'0') as i32;
                while let Some('0'..='9') = it.peek() {
                    x *= 10;
                    x += (it.next().unwrap() as u8 - b'0') as i32;
                }
                res.push(Tok::Num(x));
            }
        }
        res
    }

    fn parse_tree(it: &mut Peekable<IntoIter<Tok>>) -> TreeLink {
        let mut res = None;
        it.next();
        match it.peek() {
            Some(&Tok::Num(x)) => {
                it.next();
                res = tree!(x, Self::parse_tree(it), Self::parse_tree(it))
            }
            Some(Tok::Op(')')) => {}
            _ => panic!(),
        }
        it.next();
        res
    }
}

#[test]
fn test() {
    let codec = Codec::new();
    let root = tree!(2, tree!(1), tree!(3));
    let ans = tree!(2, tree!(1), tree!(3));
    assert_eq!(codec.deserialize(codec.serialize(root)), ans);
}

// Accepted solution for LeetCode #449: Serialize and Deserialize BST
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #449: Serialize and Deserialize BST
// /**
//  * Definition for a binary tree node.
//  * public class TreeNode {
//  *     int val;
//  *     TreeNode left;
//  *     TreeNode right;
//  *     TreeNode(int x) { val = x; }
//  * }
//  */
// public class Codec {
//     private int i;
//     private List<String> nums;
//     private final int inf = 1 << 30;
// 
//     // Encodes a tree to a single string.
//     public String serialize(TreeNode root) {
//         nums = new ArrayList<>();
//         dfs(root);
//         return String.join(" ", nums);
//     }
// 
//     // Decodes your encoded data to tree.
//     public TreeNode deserialize(String data) {
//         if (data == null || "".equals(data)) {
//             return null;
//         }
//         i = 0;
//         nums = Arrays.asList(data.split(" "));
//         return dfs(-inf, inf);
//     }
// 
//     private void dfs(TreeNode root) {
//         if (root == null) {
//             return;
//         }
//         nums.add(String.valueOf(root.val));
//         dfs(root.left);
//         dfs(root.right);
//     }
// 
//     private TreeNode dfs(int mi, int mx) {
//         if (i == nums.size()) {
//             return null;
//         }
//         int x = Integer.parseInt(nums.get(i));
//         if (x < mi || x > mx) {
//             return null;
//         }
//         TreeNode root = new TreeNode(x);
//         ++i;
//         root.left = dfs(mi, x);
//         root.right = dfs(x, mx);
//         return root;
//     }
// }
// 
// // Your Codec object will be instantiated and called as such:
// // Codec ser = new Codec();
// // Codec deser = new Codec();
// // String tree = ser.serialize(root);
// // TreeNode ans = deser.deserialize(tree);
// // return ans;

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(h)

Approach Breakdown

LEVEL ORDER

O(n) time

O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL

O(n) time

O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.