[LeetCode]Serialize and Deserialize Binary Tree 解題報告

【題目】

 

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】

其實LeetCode上樹的表示方式就挺好，即[1,2,3,null,null,4,5]這種形式，我們接下來就實現以下這種序列化。

序列化比較容易，我們做一個層次遍歷就好，空的地方用null表示，稍微不同的地方是題目中示例得到的結果是[1,2,3,null,null,4,5,null,null,null,null,]，即 4 和 5 的兩個空節點我們也存了下來。

飯序列化時，我們根據都好分割得到每個節點。需要注意的是，反序列化時如何尋找父節點與子節點的對應關系，我們知道在數組中，如果滿二叉樹（或完全二叉樹）的父節點下標是 i，那麼其左右孩子的下標分別為 2*i+1 和 2*i+2，但是這裡並不一定是滿二叉樹（或完全二叉樹），所以這個對應關系需要稍作修改。如下面這個例子：

 

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

 

序列化結果為[5,4,7,3,null,2,null,-1,null,9,null,null,null,null,null,]。

其中，節點 2 的下標是 5，可它的左孩子 9 的下標為 9，並不是 2*i+1=11，原因在於 前面有個 null 節點，這個 null 節點沒有左右孩子，所以後面的節點下標都提前了2。所以我們只需要記錄每個節點前有多少個 null 節點，就可以找出該節點的孩子在哪裡了，其左右孩子分別為 2*(i-num)+1 和 2*(i-num)+2（num為當前節點之前 null 節點的個數）。

 

【Java代碼】

 

/**
 * 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 queue = new LinkedList();
        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]; // 節點i之前null節點的個數
        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));