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.

思路

理论上说所有遍历的方法都可以。但是为了使serialize和deserialize的过程都尽量最简单，preorder是不错的选择。serialize的话，dfs比较好写，deserialize的话preorder和bfs比较好写。用“,”作为分隔符，“#”来表示null。例子里serialize之后结果就变成"1,2,3,#,#,4,5"。deserialize的时候用一个queue来保存string。

复杂度

Time: O(N), Space: O(N)

代码

// Encodes a tree to a single string.    public String serialize(TreeNode root) {        // base case        if(root == null) return "";        StringBuilder encoded = new StringBuilder();        encode(root, encoded);        return encoded.substring(1).toString();    }        private void encode(TreeNode root, StringBuilder sb) {        if(root == null) {            sb.append(",#");            return;        }        sb.append(",").append(root.val);        encode(root.left, sb);        encode(root.right, sb);    }    // Decodes your encoded data to tree.    public TreeNode deserialize(String data) {        // base case        if(data.length() == 0) return null;        Queue
    
      q = new LinkedList(Arrays.asList(data.split(",")));        return decode(q);    }        private TreeNode decode(Queue
     
       q) {        if(q.isEmpty()) return null;        String cur = q.poll();        if(cur.equals("#")) return null;        TreeNode root = new TreeNode(Integer.valueOf(cur));        root.left = decode(q);        root.right = decode(q);        return root;    }
     
    

449. Serialize and Deserialize BST

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.

The encoded string should be as compact as possible.

Note: Do not use class member/global/static variables to store states. Your serialize and deserialize algorithms should be stateless.

思路

这道题和之前不同，一般的树变成了BST，而且要求是as compact as possible。还是可以用preorder，还是需要分隔符，但是null就不需要保存了。deserialize部分要变得复杂，left的值总是小于root的值，right的值总是大于root的值，根据这个每次recursion的时候把左边的值都放到另一个queue里面，剩下的就是右边的值。

复杂度

Time: O(N^2), Space: O(N)

代码

// Encodes a tree to a single string.    public String serialize(TreeNode root) {        // base case        if(root == null) return "";        StringBuilder encoded = new StringBuilder();        encode(root, encoded);        return encoded.substring(1).toString();    }        private void encode(TreeNode root, StringBuilder sb) {        if(root == null) return;        sb.append(",").append(root.val);        encode(root.left, sb);        encode(root.right, sb);    }        // Decodes your encoded data to tree.    public TreeNode deserialize(String data) {        // base case        if(data.length() == 0) return null;        Queue
    
      q = new LinkedList();        for(String s : data.split(",")) q.offer(Integer.valueOf(s));        return decode(q);    }        private TreeNode decode(Queue
     
       q) {        if(q.isEmpty()) return null;        int cur = q.poll();        TreeNode root = new TreeNode(cur);        Queue
      
        left = new LinkedList();         while(!q.isEmpty() && q.peek() < cur) left.offer(q.poll());        root.left = decode(left);        root.right = decode(q);        return root;    }