题目
二叉树的序列化和反序列化
设计一个算法,并编写代码来序列化和反序列化二叉树。将树写入一个文件被称为“序列化”,读取文件后重建同样的二叉树被称为“反序列化”。
如何反序列化或序列化二叉树是没有限制的,你只需要确保可以将二叉树序列化为一个字符串,并且可以将字符串反序列化为原来的树结构。
样例
给出一个测试数据样例, 二叉树{3,9,20,#,#,15,7}
,表示如下的树结构:
3
/ 9 20
/ 15 7
我们的数据是进行BFS遍历得到的。当你测试结果wrong answer时,你可以作为输入调试你的代码。
你可以采用其他的方法进行序列化和反序列化。
解题
参考九章程序
看看注释就理解了,但是我表示自己想不出来Java
/** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(int val) { * this.val = val; * this.left = this.right = null; * } * } */ class Solution { /** * This method will be invoked first, you should design your own algorithm * to serialize a binary tree which denote by a root node to a string which * can be easily deserialized by your own "deserialize" method later. */ public String serialize(TreeNode root) { // write your code here if( root == null) return "{}"; ArrayList<TreeNode> queue = new ArrayList<TreeNode>(); queue.add(root); // 将二叉树的个节点按照从上到下、从左到有的存储在queue中 for(int i=0;i<queue.size();i++){ TreeNode q = queue.get(i); if(q== null) continue; queue.add(q.left); queue.add(q.right); } // 去除叶子节点的左右孩子,这个孩子是空值 while(queue.get(queue.size() - 1) == null){ queue.remove(queue.size() - 1); } // 遍历queue把转换成字符串 StringBuilder sb = new StringBuilder(); sb.append("{"); sb.append(queue.get(0).val); for(int i=1;i<queue.size(); i++){ TreeNode q = queue.get(i); if(q!= null){ sb.append(","); sb.append(q.val); }else{ sb.append(",#"); } } sb.append("}"); return sb.toString(); } /** * This method will be invoked second, the argument data is what exactly * you serialized at method "serialize", that means the data is not given by * system, it‘s given by your own serialize method. So the format of data is * designed by yourself, and deserialize it here as you serialize it in * "serialize" method. */ public TreeNode deserialize(String data) { // write your code here if(data == "{}") return null; // 以逗号分割 String[] vals = data.substring(1,data.length()-1).split(","); ArrayList<TreeNode> queue = new ArrayList<TreeNode>(); // 根节点 TreeNode root = new TreeNode(Integer.parseInt(vals[0])); queue.add(root); int index = 0; boolean isLeftChild = true; for (int i = 1; i < vals.length; i++) { if (!vals[i].equals("#")) { TreeNode node = new TreeNode(Integer.parseInt(vals[i])); if (isLeftChild) { queue.get(index).left = node; } else { queue.get(index).right = node; } queue.add(node); } if (!isLeftChild) { index++; } isLeftChild = !isLeftChild; } return root; } }
时间: 2024-12-28 14:32:35