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Java-based LeetCode algorithm problem solutions, regularly updated
package g0201_0300.s0297_serialize_and_deserialize_binary_tree;
// #Hard #String #Depth_First_Search #Breadth_First_Search #Tree #Binary_Tree #Design
// #Data_Structure_II_Day_18_Tree #Udemy_Tree_Stack_Queue
// #2022_07_06_Time_7_ms_(98.13%)_Space_51.1_MB_(74.13%)
import com_github_leetcode.TreeNode;
/*
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Codec {
private static final int BASE_OFFSET = 1000;
private static final String DELIM = "*";
private int offset;
// Encodes a tree to a single string.
public String serialize(TreeNode root) {
StringBuilder sb = new StringBuilder();
offset = 0;
serialize(root, sb);
return sb.toString();
}
public void serialize(TreeNode root, StringBuilder sb) {
// all nodes fit into 4 bits.
// IFF we offset at 0. So encode(val) = val + min(default - 1000)
if (root == null) {
sb.append(DELIM);
return;
}
String s = Integer.toHexString(root.val + BASE_OFFSET);
StringBuilder sb2 = new StringBuilder();
for (int i = 0; i < 3 - s.length(); i++) {
sb2.append('0');
}
sb2.append(s);
sb.append(sb2);
serialize(root.left, sb);
serialize(root.right, sb);
}
// Decodes your encoded data to tree.
public TreeNode deserialize(String data) {
if (data.charAt(offset) == '*') {
offset++;
return null;
}
TreeNode root =
new TreeNode(
Integer.parseInt(data.substring(offset, offset + 3), 16) - BASE_OFFSET);
offset += 3;
root.left = deserialize(data);
root.right = deserialize(data);
return root;
}
}
// Your Codec object will be instantiated and called as such:
// Codec ser = new Codec();
// Codec deser = new Codec();
// TreeNode ans = deser.deserialize(ser.serialize(root));