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.

思路:用preorder traverse,current,left, right来serialize,然后用queue来deserilize,这题跟BST不同的是,这里需要serilize NULL,因为BST有大小关系,这里没有,所以需要满树才行;而且这里就不需要Integer.MAX_VALUE和Integer.MIN_VALUE作为参数来判断是否return null了,这个复杂度是O(N);

错误点:不能想到用preorder来serialize,并且用queue来deserilize;

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Codec {
    private String NULL = "NULL";
    private String DELIMITER = "#";
    // Encodes a tree to a single string.
    public String serialize(TreeNode root) {
        StringBuilder sb = new StringBuilder();
        dfs(root, sb);
        return sb.toString();
    }
    
    private void dfs(TreeNode root, StringBuilder sb) {
        if(root == null) {
            sb.append(NULL).append(DELIMITER);
        } else {
            sb.append(root.val).append(DELIMITER);
            dfs(root.left, sb);
            dfs(root.right, sb);
        }
    }

    // Decodes your encoded data to tree.
    public TreeNode deserialize(String data) {
        if(data == null || data.equals(NULL)) {
            return null;
        }
        String[] splits = data.split(DELIMITER);
        Queue<String> queue = new LinkedList<>();
        for(String split: splits) {
            queue.offer(split);
        }
        return buildTree(queue);
    }
    
    private TreeNode buildTree(Queue<String> queue) {
        if(queue.isEmpty()) {
            return null;
        }
        String str = queue.poll();
        if(str.equals(NULL)) {
            return null;
        } else {
            TreeNode node = new TreeNode(Integer.parseInt(str));
            node.left = buildTree(queue);
            node.right = buildTree(queue);
            return node;
        }
    }
}

// Your Codec object will be instantiated and called as such:
// Codec ser = new Codec();
// Codec deser = new Codec();
// TreeNode ans = deser.deserialize(ser.serialize(root));
### 力扣热门100题列表 力扣(LeetCode)上的热门题目通常是指那些被广泛讨论、高频面试或者具有较高难度的题目。这些题目涵盖了数据结构和算法的核心知识点,适合用来提升编程能力和解决实际问题的能力。 以下是基于社区反馈整理的部分 **LeetCode Hot 100 Problems List**: #### 数组与字符串 1. Two Sum (两数之和)[^1] 2. Longest Substring Without Repeating Characters (无重复字符的最长子串)[^2] 3. Median of Two Sorted Arrays (两个有序数组的中位数)[^1] 4. Container With Most Water (盛最多水的容器)[^2] #### 链表 5. Reverse Linked List (反转链表) 6. Merge Two Sorted Lists (合并两个有序链表) 7. Remove Nth Node From End of List (删除倒数第N个节点) 8. Linked List Cycle II (环形链表II) #### 堆栈与队列 9. Valid Parentheses (有效的括号) 10. Min Stack (最小栈) 11. Sliding Window Maximum (滑动窗口最大值)[^2] #### 树与二叉树 12. Binary Tree Inorder Traversal (二叉树的中序遍历) 13. Validate Binary Search Tree (验证二叉搜索树) 14. Same Tree (相同的树) 15. Serialize and Deserialize Binary Tree (序列化与反序列化二叉树) #### 图论 16. Number of Islands (岛屿数量) 17. Course Schedule (课程表) 18. Clone Graph (克隆图) #### 排序与搜索 19. Find First and Last Position of Element in Sorted Array (在排序数组中查找元素的第一个和最后一个位置) 20. Search a 2D Matrix (二维矩阵搜索) 21. K Closest Points to Origin (最接近原点的K个点) #### 动态规划 22. Climbing Stairs (爬楼梯) 23. House Robber (打家劫舍)[^1] 24. Coin Change (零钱兑换) 25. Unique Paths (不同路径) #### 贪心算法 26. Jump Game (跳跃游戏)[^1] 27. Non-overlapping Intervals (无重叠区间) 28. Best Time to Buy and Sell Stock (买卖股票的最佳时机)[^1] #### 字符串匹配与处理 29. Implement strStr() (实现strStr()) 30. Longest Consecutive Sequence (最长连续序列) 31. Group Anagrams (分组异位词) --- ### 示例代码片段 以下是一个关于动态规划的经典例子——`Climbing Stairs` 的 Python 实现: ```python class Solution: def climbStairs(self, n: int) -> int: if n == 1 or n == 2: return n dp = [0] * (n + 1) dp[1], dp[2] = 1, 2 for i in range(3, n + 1): dp[i] = dp[i - 1] + dp[i - 2] return dp[n] ``` 上述代码通过动态规划的方式解决了 `Climbing Stairs` 问题,时间复杂度为 \(O(n)\),空间复杂度同样为 \(O(n)\)[^1]。 --- ###
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