[LeetCode] Serialize and Deserialize Binary Tree

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.

DFS

Time Complexity
serialize：O(N)
deserialize: O(N)
Space Complexity
O(N）

思路

Use preorder to serialize the tree, use "n" to denote null node and split node with ",". We can use a stringBuilder to build the string.

Use preorder to deserialize the tree, because we want to get the value from the start of the string one by one, which is like a queue. So we put the string into queue, then poll the value out as the node of the string.

代码

// Encodes a tree to a single string.
public String serialize(TreeNode root) {
    //corner case
    if(root == null) return "";
    StringBuilder sb = new StringBuilder();
    helper(root, sb);
    sb.deleteCharAt(sb.length() - 1);
    return sb.toString();
}

private void helper(TreeNode root, StringBuilder sb){
    if(root == null){
        sb.append("n").append(",");
        return;
    }
    sb.append(root.val).append(",");
    helper(root.left, sb);
    helper(root.right, sb);
}

// Decodes your encoded data to tree.
public TreeNode deserialize(String data) {
    if(data == null || data.length() == 0) return null;
    Queue queue = new LinkedList<>();
    queue.addAll(Arrays.asList(data.split(",")));
    return helper(queue);
}

private TreeNode helper(Queue queue){
    //base case
    String val = queue.poll();
    if(val.equals("n")) return null;
    TreeNode root = new TreeNode(Integer.valueOf(val));
    root.left = helper(queue);
    root.right = helper(queue);
    return root;
}