LeetCode 297. Serialize and Deserialize Binary Tree

原题链接在这里：https://leetcode.com/problems/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.

题解：

把一个树serialize 和 deserialize. 可以通过pre-order 或者 level order traversal.

用pre-order:

    1
   / \
  2   3
     / \
    4   5

例子就变成了"1,2,#,#,3,4,#,#,5,#,#". 然后deserialize先读入一个字符作为根节点，然后根节点的左右子节点递归调用deserializeHelper函数.

deserialize 时按"," 拆开，遇到"#"就是 null节点.

Note: LinkedList AddAll only works for Collection<>, array is not Collection<>. Thus it needs to use Arrays.asList(arr).

Time Complexity: serialize, O(n). deserialize, O(n). Space: O(n).

AC Java:

/**
 * 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();
        preorder(root, sb);
        return sb.toString();
    }
    private void preorder(TreeNode root, StringBuilder sb){
        if(root == null){
            sb.append("#").append(",");
            return;
        }
        sb.append(root.val).append(",");
        preorder(root.left, sb);
        preorder(root.right, sb);
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        LinkedList<String> que = new LinkedList<String>();
        que.addAll(Arrays.asList(data.split(",")));
        return deserial(que);
    }
    
    private TreeNode deserial(LinkedList<String> que){
        String str = que.pollFirst();
        if(str.equals("#")){
            return null;
        }
        TreeNode root = new TreeNode(Integer.valueOf(str));
        root.left = deserial(que);
        root.right = deserial(que);
        return root;
    }
}

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

Iteration 方法使用的BFS. 当处理serialize, queue不为空时，弹出节点cur, 若cur是null, 就在sb后面加上"#,"; 若不是null, 就在sb后面加上cur.val 和 ",".

处理deserialize时，先把string 用"," 格成string 数组。 如果是满树，那么左节点index就是2*i+1, 右节点index是2*i+2. 但这里不是满树。

需要同时记录当前i前面有几个null节点，用count存储，左节点index变成2*(i-count[i]) + 1, 右节点index变成2*(i-count[i]) + 2.

Note: 记住Queue 是 abstract class, 不能直接用来生成 que, 需要用LinkedList.

AC Java:

public class Codec {

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

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        String [] s = data.split(",");
        int [] count = new int[s.length];
        TreeNode [] tns = new TreeNode[s.length];
        
        for(int i = 0; i<s.length; i++){
            if(i>0){
                count[i] = count[i-1];
            }
            if(s[i].equals("#")){
                tns[i] = null;
                count[i]++;
            }else{
                tns[i] = new TreeNode(Integer.valueOf(s[i]));
            }
        }
        
        for(int i = 0; i<s.length; i++){
            if(s[i].equals("#")){
                continue;
            }
            tns[i].left = tns[2 * (i-count[i]) + 1];
            tns[i].right = tns[2 * (i-count[i]) + 2];
        }
        return tns[0];
    }
}

类似Serialize and Deserialize BST, Serialize and Deserialize N-ary Tree, Encode and Decode Strings, Find Duplicate Subtrees.

Reference: http://blog.csdn.net/ljiabin/article/details/49474445