# Serialize and Deserialize Binary Tree

`Tree`, `Design`

Hard

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:

``````You may serialize the following tree:

1
/ \
2   3
/ \
4   5

as
"[1,2,3,null,null,4,5]"
``````

Clarification:The above format is the same ashow 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.

## Solution & Analysis

Pre-Oder Traversal:

The preorder DFS traverse follows recursively the order of root -> left subtree -> right subtree.

``````[1,2,3,null,null,4,5]

---serialize--->

1,2,#,#,3,4,#,#,5,#,#,
``````

Complexity Analysis

• Time complexity : in both serialization and deserialization functions, we visit each node exactly once, thus the time complexity is O(N), whereNNis the number of nodes, _i.e._the size of tree.

• Space complexity : in both serialization and deserialization functions, we keep the entire tree, either at the beginning or at the end, therefore, the space complexity is O(N).

10 ms, faster than 92.02%

``````/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Codec {

// Encodes a tree to a single string.
public String serialize(TreeNode root) {
if (root == null) {
return "";
}
StringBuilder sb = new StringBuilder();

traverse(root, sb);
// System.out.println(sb.toString());

return sb.toString();
}

private void traverse(TreeNode root, StringBuilder sb) {
if (root != null) {
sb.append(String.valueOf(root.val));
sb.append(",");
traverse(root.left, sb);
traverse(root.right, sb);
} else {
sb.append("#");
sb.append(",");
}
}

// Decodes your encoded data to tree.
public TreeNode deserialize(String data) {
if (data == null || data.isEmpty()) {
return null;
}
String[] values = data.split(",");

return buildTree(queue);
}

private TreeNode buildTree(Queue<String> queue) {
if (queue == null || queue.isEmpty()) {
return null;
}
String value = queue.poll();
if (value.equals("#")) {
return null;
}

TreeNode node = new TreeNode(Integer.parseInt(value));
node.left = buildTree(queue);
node.right = buildTree(queue);
return node;
}
}

// Your Codec object will be instantiated and called as such:
// Codec codec = new Codec();
// codec.deserialize(codec.serialize(root));
``````

via @cdai -- 以下：代码比较简洁，但是serial的部分不是很readable的地方在于，何时加delimiter `","`

``````1,2,#,#,3,4,#,#,5,#,#
``````

``````/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Codec {

public String serialize(TreeNode root) {
return serial(new StringBuilder(), root).toString();
}

// Generate preorder string
private StringBuilder serial(StringBuilder str, TreeNode root) {
if (root == null) return str.append("#");
str.append(root.val).append(",");
serial(str, root.left).append(",");
serial(str, root.right);
return str;
}

public TreeNode deserialize(String data) {
}

// Use queue to simplify position move
private TreeNode deserial(Queue<String> q) {
String val = q.poll();
if ("#".equals(val)) return null;
TreeNode root = new TreeNode(Integer.valueOf(val));
root.left = deserial(q);
root.right = deserial(q);
return root;
}
}
``````

Serialize时类似 pre-order traversal. 这里为了节省空间，删去了末尾的"#"，虽然deserialized的时候并无影响。

Deserialize的时候要从左边节点开始填充，也就是说只有i % 2 == 1时才从queue中poll()，如果不为null，再生成新节点放入queue中。

``````/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Codec {

// Encodes a tree to a single string.
public String serialize(TreeNode root) {
if (root == null) {
return "";
}
StringBuilder sb = new StringBuilder();
// sb.append('[');

queue.offer(root);

while (!queue.isEmpty()) {
TreeNode node = queue.poll();
if (node == null) {
sb.append('#');
} else {
sb.append(node.val);
queue.offer(node.left);
queue.offer(node.right);
}
sb.append(',');
}

// (optional, won't affect deserialization) remove tailing sharp and comma
for (int i = sb.length() - 1; i >= 0; i--) {
if (sb.charAt(i) == '#' || sb.charAt(i) == ',') {
sb.deleteCharAt(i);
} else {
break;
}
}

// sb.append(']');
return sb.toString();
}

// Decodes your encoded data to tree.
public TreeNode deserialize(String data) {
if (data == null || data.isEmpty()) {
return null;
}
String[] values = data.split(",");
TreeNode root = new TreeNode(Integer.parseInt(values[0]));
q.offer(root);

int i = 1;
boolean isLeft = true;
TreeNode node = null;
while (i < values.length) {
if (isLeft) {
node = q.poll();
if (values[i].equals("#")) {
node.left = null;
} else {
node.left = new TreeNode(Integer.parseInt(values[i]));
q.offer(node.left);
}
} else {
if (values[i].equals("#")) {
node.right = null;
} else {
node.right = new TreeNode(Integer.parseInt(values[i]));
q.offer(node.right);
}
}
isLeft = !isLeft;
i++;
}
return root;
}
}

// Your Codec object will be instantiated and called as such:
// Codec codec = new Codec();
// codec.deserialize(codec.serialize(root));
``````

## Reference

https://leetcode.com/problems/serialize-and-deserialize-binary-tree/solution/

https://leetcode.com/problems/serialize-and-deserialize-binary-tree/discuss/74253/Easy-to-understand-Java-Solution

https://www.jiuzhang.com/solutions/binary-tree-serialization/#tag-other

https://leetcode.com/problems/serialize-and-deserialize-binary-tree/discuss/74260/Recursive-DFS-Iterative-DFS-and-BFS