【LeetCode】Serialize and Deserialize Binary Tree 解题报告

【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.

【解析】

其实LeetCode上树的表示方式就挺好，即"[1,2,3,null,null,4,5]"这种形式，我们接下来就实现以下这种序列化。

序列化比较容易，我们做一个层次遍历就好，空的地方用null表示，稍微不同的地方是题目中示例得到的结果是"[1,2,3,null,null,4,5,null,null,null,null,]"，即 4 和 5 的两个空节点我们也存了下来。

饭序列化时，我们根据都好分割得到每个节点。需要注意的是，反序列化时如何寻找父节点与子节点的对应关系，我们知道在数组中，如果满二叉树（或完全二叉树）的父节点下标是 i，那么其左右孩子的下标分别为 2*i+1 和 2*i+2，但是这里并不一定是满二叉树（或完全二叉树），所以这个对应关系需要稍作修改。如下面这个例子：

5
      / \
     4   7
    /   /
   3   2
  /   /
 -1  9      

序列化结果为[5,4,7,3,null,2,null,-1,null,9,null,null,null,null,null,]。

其中，节点 2 的下标是 5，可它的左孩子 9 的下标为 9，并不是 2*i+1=11，原因在于 前面有个 null 节点，这个 null 节点没有左右孩子，所以后面的节点下标都提前了2。所以我们只需要记录每个节点前有多少个 null 节点，就可以找出该节点的孩子在哪里了，其左右孩子分别为 2*(i-num)+1 和 2*(i-num)+2（num为当前节点之前 null 节点的个数）。

【Java代码】

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Codec {

    // Encodes a tree to a single string.
    public String serialize(TreeNode root) {
        StringBuilder sb = new StringBuilder();
        
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        
        while (!queue.isEmpty()) {
            TreeNode node = queue.poll();
            if (node == null) {
                sb.append("null,");
            } else {
                sb.append(String.valueOf(node.val) + ",");
                queue.offer(node.left);
                queue.offer(node.right);
            }
        }
        
        return sb.toString();
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        if (data.isEmpty()) return null;
        
        String[] vals = data.split(",");
        int[] nums = new int[vals.length]; // 节点i之前null节点的个数
        TreeNode[] nodes = new TreeNode[vals.length];
        
        for (int i = 0; i < vals.length; i++) {
            if (i > 0) {
                nums[i] = nums[i - 1];
            }
            if (vals[i].equals("null")) {
                nodes[i] = null;
                nums[i]++;
            } else {
                nodes[i] = new TreeNode(Integer.parseInt(vals[i]));
            }
        }
        
        for (int i = 0; i < vals.length; i++) {
            if (nodes[i] == null) {
                continue;
            }
            nodes[i].left = nodes[2 * (i - nums[i]) + 1];
            nodes[i].right = nodes[2 * (i - nums[i]) + 2];
        }
        
        return nodes[0];
    }
}

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

这道题比较简洁的写法是递归，参见https://leetcode.com/discuss/66117/easy-to-understand-java-solution。