LeetCode OJ 297. Serialize and Deserialize Binary Tree

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.

【解析1】

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】非递归

/**
 * 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 == null || 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++) {   //计算每个节点前面null节点的数目
            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++) {  //对节点进行连接操作
            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];
    }

}

 

【解析2】

我们也可以用递归来解决这个问题：The idea is simple: print the tree in pre-order traversal and use "X" to denote null node and split node with ",". We can use a StringBuilder for building the string on the fly. For deserializing, we use a Queue to store the pre-order traversal and since we have "X" as null node, we know exactly how to where to end building subtress.

这个思路用到的是树的前序遍历，因为序列中包含了值为null的节点，因此我们可以很容易地进行反序列化操作。

【java代码】递归

public class Codec {
    private static final String spliter = ",";
    private static final String NN = "X";

    // Encodes a tree to a single string.
    public String serialize(TreeNode root) {
        StringBuilder sb = new StringBuilder();
        buildString(root, sb);
        return sb.toString();
    }

    private void buildString(TreeNode node, StringBuilder sb) {
        if (node == null) {
            sb.append(NN).append(spliter);
        } else {
            sb.append(node.val).append(spliter);
            buildString(node.left, sb);
            buildString(node.right,sb);
        }
    }
    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        Deque<String> nodes = new LinkedList<>();
        nodes.addAll(Arrays.asList(data.split(spliter)));
        return buildTree(nodes);
    }

    private TreeNode buildTree(Deque<String> nodes) {
        String val = nodes.remove();
        if (val.equals(NN)) return null;
        else {
            TreeNode node = new TreeNode(Integer.valueOf(val));
            node.left = buildTree(nodes);
            node.right = buildTree(nodes);
            return node;
        }
    }
}