LeetCode之二叉树的序列化与反序列化

题目链接：https://leetcode-cn.com/problems/serialize-and-deserialize-binary-tree/

解题步骤：

• 通过前序遍历将二叉树序列化为字符串，前序遍历顺序：根节点、左子树、右子树
• 节点间用","隔开，并用"null"表示空节点
• 使用递归，我们可以得到序列：“1,2,null,null,3,4,null,null,5,null,null,”
• 获得序列化字符串后，第一个值为根节点，接下来是左子树节点，右子树节点
• 所以我们可以依次取到节点的值，并使用前序遍历的方式，构建二叉树

C++实现：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Codec {
public:
    void rserialize(TreeNode* root, string &str) {
        if (root == nullptr)
            str += "null,";
        else {
            str += (to_string(root->val) + ",");
            rserialize(root->left, str);
            rserialize(root->right, str);
        }
    }

    // Encodes a tree to a single string.
    string serialize(TreeNode* root) {
        string str = "";
        rserialize(root, str);
        return str;
    }

    TreeNode* rdeserialize(string &data, int &index) {
        if (data[index] == 'n') {  // 跳过null节点
            index += 5;
            return NULL;
        }
        int value = 0;
        int flag = 1;
        while (data[index] != ',') {
            if (data[index] == '-'){  // 判断是否为负数
                flag = -1;
                index++;
            }
            value = 10 * value + (data[index] - '0');
            index++;
        }
        index++;
        TreeNode* root = new TreeNode(flag * value);  // 根节点
        root->left = rdeserialize(data, index);  // 构建左子树
        root->right = rdeserialize(data, index);  // 构建右子树

        return root;
    }

    // Decodes your encoded data to tree.
    TreeNode* deserialize(string data) {
        int index = 0;
        return rdeserialize(data, index);
    }
};

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