297. Serialize and Deserialize Binary Tree

题目：

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.

链接： http://leetcode.com/problems/serialize-and-deserialize-binary-tree/

5/14/2017

算法班，deserialize部分抄的

层序遍历的方法

注意

1. 第36行需要减去最后一个","

2. 第44行将string里面的各个char转为数字和#

3. deserialize部分，TreeNode部分来自于queue，新建的子树node值来自于values

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Codec {
    private final String nullNode = "#";
    private final String delimiter = ",";
    
    // Encodes a tree to a single string.
    public String serialize(TreeNode root) {
        StringBuilder sb = new StringBuilder();
        if (root == null) return "#";

        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        sb.append("[");
        queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode node = queue.poll();
                if (node == null) {
                    sb.append(nullNode);
                } else {
                    sb.append(node.val);
                    queue.offer(node.left);
                    queue.offer(node.right);
                }
                sb.append(delimiter);
            }
        }
        sb.deleteCharAt(sb.length() - 1);
        sb.append("]");
        return sb.toString();
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        if (data.equals("#")) return null;
        String[] values = data.substring(1, data.length() - 1).split(delimiter);
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        TreeNode root = new TreeNode(Integer.parseInt(values[0]));
        
        queue.offer(root);
        for (int i = 1; i < values.length; i++) {
            // node comes from queue, left/right comes from values
            TreeNode parent = queue.poll();
            if (!values[i].equals(nullNode)) {
                TreeNode left = new TreeNode(Integer.parseInt(values[i]));
                parent.left = left;
                queue.add(left);
            }
            i++;
            if (!values[i].equals(nullNode)) {
                TreeNode right = new TreeNode(Integer.parseInt(values[i]));
                parent.right = right;
                queue.add(right);
            }
        }
        return root;
    }
}

// Your Codec object will be instantiated and called as such:
// Codec codec = new Codec();
// codec.deserialize(codec.serialize(root));

也可以递归调用

http://www.cnblogs.com/yrbbest/p/5047035.html

别人的答案

https://discuss.leetcode.com/topic/30195/recursive-dfs-iterative-dfs-and-bfs

更多讨论

https://discuss.leetcode.com/category/375/serialize-and-deserialize-binary-tree