# 297. Serialize and Deserialize Binary Tree
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]"
# Solution
Approach 1: level order.
Approach 2: pre-order.
# Code (Python)
Approach 1:
from collections import deque
NULL_NODE = '#'
class Codec:
def serialize0(self, root):
# level order
queue = deque([root])
vals = []
while queue:
node = queue.popleft()
if not node:
vals.append(NULL_NODE)
else:
vals.append(str(node.val))
queue.append(node.left)
queue.append(node.right)
return ','.join(vals)
def deserialize(self, data):
# level order
# a good place to use an iterator!
vals = iter(data.split(','))
root_val = next(vals)
if root_val == NULL_NODE:
return None
root = TreeNode(int(root_val))
queue = deque([root])
while True:
left_val, right_val = next(vals, None), next(vals, None) # provide a default value as an alternative to StopIteration error
if not left_val or not right_val:
return root
parent = queue.popleft()
parent.left = TreeNode(int(left_val)) if left_val != NULL_NODE else None
parent.right = TreeNode(int(right_val)) if right_val != NULL_NODE else None
for node in (parent.left, parent.right):
if node:
queue.append(node)
Approach 2:
def serialize(self, root):
# preorder
vals = []
def recurse(node):
if not node:
vals.append(NULL_NODE)
return
vals.append(str(node.val))
recurse(node.left)
recurse(node.right)
recurse(root)
return ','.join(vals)
def deserialize(self, data):
# preorder
vals = iter(data.split(','))
def recurse():
next_value = next(vals)
if next_value == NULL_NODE:
return None
node = TreeNode(int(next_value))
node.left = recurse() # won't exhaust the iterator because nodes are serialized the same way -- should be just right
node.right = recurse()
return node
return recurse()
'''
#OR:
splitted = data.split(',')
def explore(index):
# returns (node, index_to_consume_next)
if index >= len(splitted) or splitted[index] == NULL_NODE:
return (None, index + 1)
root = TreeNode(int(splitted[index]))
root.left, new_index = explore(index + 1)
root.right, end_index = explore(new_index)
return (root, end_index)
return explore(0)[0]
'''
# Code (C++)
Approach 1:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
// Iteration - level by level.
class Codec {
public:
// Encodes a tree to a single string.
string serialize(TreeNode* root) {
queue<TreeNode*> q;
if (root)
q.push(root);
ostringstream oss;
while (!q.empty())
{
int qSize = q.size();
for (int i = 0; i < qSize; ++i)
{
TreeNode *node = q.front();
q.pop();
if (node)
oss << node->val << ",";
else
oss << "null,";
if (node)
{
q.push(node->left);
q.push(node->right);
}
}
}
return oss.str();
}
// Decodes your encoded data to tree.
TreeNode* deserialize(string data) {
queue<string> q1;
int head = 0;
for (int tail = 0; tail < data.size(); ++tail)
{
if (data[tail] == ',')
{
q1.push(data.substr(head, tail-head));
head = tail + 1;
}
}
TreeNode *root = NULL;
if (!q1.empty())
{
root = new TreeNode(stoi(q1.front()));
q1.pop();
}
queue<TreeNode*> q2;
q2.push(root);
while (!q1.empty())
{
TreeNode *node = q2.front();
q2.pop();
// left child
string left = q1.front();
q1.pop();
if (left != "null")
{
node->left = new TreeNode(stoi(left));
q2.push(node->left);
}
// right child
string right = q1.front();
q1.pop();
if (right != "null")
{
node->right = new TreeNode(stoi(right));
q2.push(node->right);
}
}
return root;
}
};
// Your Codec object will be instantiated and called as such:
// Codec codec;
// codec.deserialize(codec.serialize(root));
Approach 2:
// Recursion - preorder traversal.
class Codec {
private:
TreeNode* doDeserialize(istringstream& iss) {
string str;
iss >> str;
if (str == "N")
return NULL;
TreeNode *node = new TreeNode(stoi(str));
node->left = doDeserialize(iss);
node->right = doDeserialize(iss);
return node;
}
public:
// Encodes a tree to a single string.
string serialize(TreeNode* root) {
if (root == NULL)
return "N ";
ostringstream oss;
oss << root->val << ' '
<< serialize(root->left)
<< serialize(root->right);
return oss.str();
}
// Decodes your encoded data to tree.
TreeNode* deserialize(string data) {
istringstream iss(data);
return doDeserialize(iss);
}
};