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October 31, 2021 10:46
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import java.util.LinkedList; | |
import java.util.Queue; | |
public class SerializeDeserializeBinaryTree { | |
//Definition for a binary tree node. | |
public class TreeNode { | |
int val; | |
TreeNode left; | |
TreeNode right; | |
TreeNode(int x) { val = x; } | |
} | |
public String serialize(TreeNode root) { //Serialization: takes a tree node, returns a string representing the node | |
if (root==null) return ""; | |
Queue<TreeNode> qu=new LinkedList<>(); //Using queue for level-oder traversal of binary tree (BFS) | |
StringBuilder sb=new StringBuilder(); | |
qu.offer(root); | |
sb.append(String.valueOf(root.val)); | |
sb.append(' '); //Nodes separated by ' ' character | |
while (!qu.isEmpty()) { | |
TreeNode x=qu.poll(); | |
if (x.left==null) sb.append("null "); // Handling nodes with no left child | |
else { | |
qu.offer(x.left); | |
sb.append(String.valueOf(x.left.val)); // Left node ID is added first to the string | |
sb.append(' '); | |
} | |
if (x.right==null) sb.append("null "); // Handling nodes with no right child | |
else { | |
qu.offer(x.right); | |
sb.append(String.valueOf(x.right.val)); // Right node ID is added last to the string | |
sb.append(' '); | |
} | |
} | |
return sb.toString(); | |
} | |
public TreeNode deserialize(String data) { //Deserialization: takes a string, returns the root node of the tree it represents | |
if (data.length()==0) return null; | |
String[] node=data.split(" "); //Split the string to a string array for easier iteration | |
Queue<TreeNode> qu=new LinkedList<>(); | |
TreeNode root=new TreeNode(Integer.valueOf(node[0])); //Make tree root with '0' as its identifier | |
qu.offer(root); | |
int i=1; | |
// For each node in the queue, its left and right children are assigned based on the level-order traversal | |
while (!qu.isEmpty()) { | |
TreeNode x=qu.poll(); | |
if (node[i].equals("null")) x.left=null; // Handling nodes with no left child | |
else { | |
x.left=new TreeNode(Integer.valueOf(node[i])); | |
qu.offer(x.left); //Child node is added to queue for further traversal of tree. Left node should be first. | |
} | |
i++; | |
if (node[i].equals("null")) x.right=null; // Handling nodes with no right child | |
else { | |
x.right=new TreeNode(Integer.valueOf(node[i])); | |
qu.offer(x.right); //Child node is added to queue for further traversal of tree. Right node should be last. | |
} | |
i++; | |
} | |
return root; | |
} | |
} |
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