序列化是将一个数据结构或者对象转换为连续的比特位的操作，进而可以将转换后的数据存储在一个文件或者内存中，同时也可以通过网络传输到另一个计算机环境，采取相反方式重构得到原数据。 请设计一个算法来实现二叉树的序列化与反序列化。这里不限定你的序列 / 反序列化算法执行逻辑，你只需要保证一个二叉树可以被序列化为一个字符串并且将这个字符串反序列化为原始的树结构。
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ BFS // 序列化为“1,2,3,null,null,4,5,null,null,null,null,” public class Codec { // Encodes a tree to a single string. public String serialize(TreeNode root) { String str = ""; Queue<TreeNode> queue = new LinkedList<TreeNode>(); queue.offer(root); while (!queue.isEmpty()) { TreeNode node = queue.poll(); if (node != null) { str += String.valueOf(node.val) + ","; queue.offer(node.left); queue.offer(node.right); } else { str += "null,"; } } return str; } // Decodes your encoded data to tree. public TreeNode deserialize(String data) { String[] data_array = data.split(","); int i = 0, n = data_array.length; if (data_array[i].equals("null")) { return null; } Queue<TreeNode> queue = new LinkedList<TreeNode>(); TreeNode root = new TreeNode(Integer.valueOf(data_array[i++])); queue.offer(root); while (!queue.isEmpty()) { TreeNode node = queue.poll(); if (data_array[i].equals("null")) { node.left = null; } else { TreeNode lnode = new TreeNode(Integer.valueOf(data_array[i])); node.left = lnode; queue.offer(lnode); } i++; if (data_array[i].equals("null")) { node.right = null; } else { TreeNode rnode = new TreeNode(Integer.valueOf(data_array[i])); node.right = rnode; queue.offer(rnode); } i++; } return root; } } DFS // 序列化为"1,2,3,null,null,4,null,null,5,null,null," public class Codec { public String rserialize(TreeNode root, String str) { if (root == null) { str += "null,"; } else { str += String.valueOf(root.val) + ","; str = rserialize(root.left, str); str = rserialize(root.right, str); } return str; } // Encodes a tree to a single string. public String serialize(TreeNode root) { return rserialize(root, ""); } public TreeNode rdeserialize(List<String> l) { if (l.get(0).equals("null")) { l.remove(0); return null; } TreeNode root = new TreeNode(Integer.valueOf(l.get(0))); l.remove(0); root.left = rdeserialize(l); root.right = rdeserialize(l); return root; } // Decodes your encoded data to tree. public TreeNode deserialize(String data) { String[] data_array = data.split(","); List<String> data_list = new LinkedList<String>(Arrays.asList(data_array)); return rdeserialize(data_list); } } 括号表示编码 + 递归下降解码 转载自官解： 作者：LeetCode-Solution 链接：https://leetcode-cn.com/problems/serialize-and-deserialize-binary-tree/solution/er-cha-shu-de-xu-lie-hua-yu-fan-xu-lie-hua-by-le-2/ public class Codec { public String serialize(TreeNode root) { if (root == null) { return "X"; } String l = "(" + serialize(root.left) + ")"; String r = "(" + serialize(root.right) + ")"; return l + root.val + r; } public TreeNode deserialize(String data) { int[] ptr = {0}; return parse(data, ptr); } public TreeNode parse(String data, int[] ptr) { if (data.charAt(ptr[0]) == 'X') { ++ptr[0]; return null; } TreeNode cur = new TreeNode(0); cur.left = parseSubtree(data, ptr); cur.val = parseInt(data, ptr); cur.right = parseSubtree(data, ptr); return cur; } public TreeNode parseSubtree(String data, int[] ptr) { ++ptr[0]; // 跳过左括号 TreeNode subtree = parse(data, ptr); ++ptr[0]; // 跳过右括号 return subtree; } public int parseInt(String data, int[] ptr) { int x = 0, sgn = 1; if (!Character.isDigit(data.charAt(ptr[0]))) { sgn = -1; ++ptr[0]; } while (Character.isDigit(data.charAt(ptr[0]))) { x = x * 10 + data.charAt(ptr[0]++) - '0'; } return x * sgn; } }