Reference: LeetCode
Difficulty: Hard

Problem

Serialization is the process of 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 tree. There is no restriction on how your serialization/deserialization algorithm should work. You just need to ensure that a binary tree can be serialized to a string and this string can be deserialized to the original tree structure.

Example:

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You may serialize the following tree:
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/ \
2 3
/ \
4 5
as "[1,2,3,null,null,4,5]"

Clarification: The above format is the same as how LeetCode serializes a binary tree. You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.

Note: Do not use class member/global/static variables to store states. Your serialize and deserialize algorithms should be stateless.

Analysis

BFS

Note:

  • Check the null case.
  • Handle , in the string. Remove it in the end.
  • Use StringBuffer.
  • Compatible with null nodes.

Time: $O(N)$ if using StringBuffer for string concatenation.
Space: $O(N)$ since we need to keep the entire tree.

Serialize:

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    1
/ \
2 3
/ \
4 5
Result: "1,2,3,null,null,4,5,null,null,null,null"
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// Encodes a tree to a single string.
public String serialize(TreeNode root) {
if (root == null) {
return null;
}
StringBuffer sb = new StringBuffer();
serializeHelper(root, sb);
sb.setLength(sb.length() - 1); // "remove the last ','"

return sb.toString();
}

private void serializeHelper(TreeNode root, StringBuffer sb) {
if (root == null) {
return;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);

while (queue.size() > 0) {
int childSize = queue.size();
for (int i = 0; i < childSize; ++i) {
TreeNode p = queue.poll();
if (p == null) {
sb.append("null,");
} else {
sb.append(p.val + ",");
queue.offer(p.left);
queue.offer(p.right);
}
}
}
}

Deserialize:

  • Handle the base cases.
  • Before the while loop, handle the root node first. So you don’t need an extra dummy node.
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// Decodes your encoded data to tree.
public TreeNode deserialize(String data) {
if (data == null || data.length() == 0) {
return null;
}
List<Integer> list = parseString(data);
Queue<TreeNode> queue = new LinkedList<>();
int index = 0;
TreeNode root = new TreeNode(list.get(index));
queue.offer(root);
index = 1;
while (queue.size() > 0) {
int childSize = queue.size();
for (int i = 0; i < childSize; ++i) {
TreeNode curr = queue.poll();
if (curr == null) continue;
// left
Integer val = list.get(index);
curr.left = (val != null) ? new TreeNode(val) : null; // can omit index checking
index += 1;
// right
val = list.get(index);
curr.right = (val != null) ? new TreeNode(val) : null;
index += 1;
queue.offer(curr.left); // may be null
queue.offer(curr.right);
}
}
return root;
}

private List<Integer> parseString(String data) {
String[] parsedStrings = data.split(",");
List<Integer> result = new ArrayList<>();
for (String s : parsedStrings) {
result.add((s.equals("null")) ? null : Integer.parseInt(s));
}
return result;
}

DFS (Solution)

  • We don’t need to remove the last ,. The return list after split does not include "".
  • If root is null, the string is null.

Time: $O(N)$ if using StringBuffer for string concatenation.
Space: $O(N)$ since we need to keep the entire tree.

Serialize:

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

private void rserialize(TreeNode root, StringBuffer sb) {
if (root == null) {
sb.append("null,");
return; // very important
}
sb.append(root.val + ",");
rserialize(root.left, sb);
rserialize(root.right, sb);
}

Deserialize:

  • In rdeserialize, null check for L is not necessary. We assume the input is a valid preorder input.
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public TreeNode deserialize(String data) {
String[] dataArray = data.split(",");
List<String> dataList = new LinkedList<>(Arrays.asList(dataArray));
return rdeserialize(dataList);
}

private TreeNode rdeserialize(List<String> L) {
if (L.get(0).equals("null")) {
L.remove(0);
return null;
}
TreeNode root = new TreeNode(Integer.valueOf(L.get(0)));
L.remove(0);
root.left = rdeserialize(L);
root.right = rdeserialize(L);

return root;
}

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