题目描述

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.

For example, you may serialize the following tree

    1/ \2   3/ \4   5

as “[1,2,3,null,null,4,5]”, just the same as how LeetCode OJ serializes a binary tree. You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.

Note: Do not use class member/global/static variables to store states. Your serialize and deserialize algorithms should be stateless.

Credits:
Special thanks to @Louis1992 for adding this problem and creating all test cases.

分析

我的最开始的思路是：先求树的深度h，把树中的每个结点（2^h-1）个结点，全部记录下来，即使是null。但是这样有些极端的测试用例会超时，比如树中每个结点只有右结点，树高1000。用这种方法要遍历2^h-1次，显然能够优化。

参考 how LeetCode OJ serializes a binary tree的序列化方式，下面的二叉树，序列化后的String可以是”1,2,3,null,null,4,null,5,null”，这种方法在序列化二叉树时，只用树结点数量规模的字符即可，省时省空间。

    1/ \2   3/ 4   /5

代码

第一种方法：超时的代码

    // Encodes a tree to a single string.public static String serialize(TreeNode root) {if (root == null) {return "";}StringBuffer sb = new StringBuffer();Deque<TreeNode> deque = new LinkedList<TreeNode>();deque.add(root);int n = (int) (Math.pow(2, maxDepth(root)) - 1);while (n-- != 0) {TreeNode p = deque.pop();if (p == null) {sb.append(",#");deque.add(null);deque.add(null);} else {sb.append("," + p.val);deque.add(p.left);deque.add(p.right);}}return sb.toString().substring(1);}// Decodes your encoded data to tree.public static TreeNode deserialize(String data) {if (data == null || data.length() == 0) {return null;}String[] s = data.split(",");TreeNode[] nodeArray = new TreeNode[s.length];for (int i = 0; i < nodeArray.length; i++) {if (!"#".equals(s[i])) {nodeArray[i] = new TreeNode(Integer.valueOf(s[i]));}}for (int parent = 0; parent < s.length / 2; parent++) {if (nodeArray[parent] == null) {continue;}nodeArray[parent].left = nodeArray[parent * 2 + 1];nodeArray[parent].right = nodeArray[parent * 2 + 2];}return nodeArray[0];}public static int maxDepth(TreeNode root) {if (root == null) {return 0;}int nLeft = maxDepth(root.left);int nRight = maxDepth(root.right);return nLeft > nRight ? (nLeft + 1) : (nRight + 1);}

第二种方法，AC的代码

    public String serialize(TreeNode root) {if (root == null) {return "";}StringBuffer sb = new StringBuffer();Deque<TreeNode> deque = new LinkedList<TreeNode>();deque.add(root);while (!deque.isEmpty()) {TreeNode p = deque.pop();if (p == null) {sb.append(",#");} else {sb.append("," + p.val);deque.add(p.left);deque.add(p.right);}}// 第一个元素前也有一个逗号，截取return sb.toString().substring(1);}public TreeNode deserialize(String data) {if (data == null || data.length() == 0) {return null;}String[] s = data.split(",");TreeNode[] node = new TreeNode[s.length];// 新建TreeNode，并初始化for (int i = 0; i < node.length; i++) {if (!"#".equals(s[i])) {node[i] = new TreeNode(Integer.valueOf(s[i]));}}int parent = 0;// 将结点连接起来for (int i = 0; parent * 2 + 2 < s.length; i++) {if (node[i] != null) {node[i].left = node[parent * 2 + 1];node[i].right = node[parent * 2 + 2];parent++;}}return node[0];}

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题目描述

分析

代码

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