• 0297. Serialize and Deserialize Binary Tree (H)


    Serialize and Deserialize Binary Tree (H)

    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.

    Clarification: The input/output format is the same as how LeetCode serializes a binary tree. You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.

    Example 1:

    Input: root = [1,2,3,null,null,4,5]
    Output: [1,2,3,null,null,4,5]
    

    Example 2:

    Input: root = []
    Output: []
    

    Example 3:

    Input: root = [1]
    Output: [1]
    

    Example 4:

    Input: root = [1,2]
    Output: [1,2]
    

    Constraints:

    • The number of nodes in the tree is in the range [0, 104].
    • -1000 <= Node.val <= 1000

    题意

    实现二叉树的序列化和反序列化。

    思路

    按照前序遍历的顺序序列化二叉树,形式为结点的值用分隔符相连。反序列化同样参照前序遍历的顺序,递归处理即可。


    代码实现

    Java

    public class Codec {
        // Encodes a tree to a single string.
        public String serialize(TreeNode root) {
            if (root == null) {
                return "null";
            }
    
            String s = "" + root.val;
            s +="," + serialize(root.left);
            s += "," + serialize(root.right);
    
            return s;
        }
    
        // Decodes your encoded data to tree.
        public TreeNode deserialize(String data) {
            String[] arr = data.split(",");
            Queue<TreeNode> q = new LinkedList<>();
            for (int i = 0; i < arr.length; i++) {
                q.offer(arr[i].equals("null") ? null : new TreeNode(Integer.parseInt(arr[i])));
            }
            return dfs(q);
        }
    
        private TreeNode dfs(Queue<TreeNode> q) {
            if (q.isEmpty()) {
                return null;
            }
    
            TreeNode cur = q.poll();
            if (cur == null) return null;
            cur.left = dfs(q);
            cur.right = dfs(q);
            return cur;
        }
    }
    
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  • 原文地址:https://www.cnblogs.com/mapoos/p/15208041.html
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