LC-297-二叉树的序列化与反序列化

原题链接：二叉树的序列化与反序列化

题解

DFS

思路：

前序遍历二叉树，遇到空则将空表示为null，然后构造的时候再按照字符串进行前序构造。

时间复杂度：$O(n)$

代码：

/**
 * 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 preOrder(TreeNode *root, string &s) {
        if(root == NULL) {
            s += " null";
            return;
        }
        s += " " + to_string(root->val);
        preOrder(root->left, s);
        preOrder(root->right, s);
    }

    vector<string> split(string &s, char delim) {
        stringstream ss(s);
        string item;
        vector<string> res;
        while(getline(ss, item, delim)) {
            if(!item.empty())
                res.push_back(item);
        }
        return res;
    }

    // Encodes a tree to a single string.
    string serialize(TreeNode* root) {
        string res;
        preOrder(root, res);
        return res;
    }

    int cur = 0;
    TreeNode *build(vector<string> &v) {
        if(v[cur] == "null") {
            cur ++;
            return NULL;
        }
        TreeNode *root = new TreeNode(atoi(v[cur ++].c_str()));
        root->left = build(v);
        root->right = build(v);
        return root;
    }

    // Decodes your encoded data to tree.
    TreeNode* deserialize(string data) {
        vector<string> v = split(data, ' ');
        cur = 0;
        return build(v);
    }
};

// Your Codec object will be instantiated and called as such:
// Codec ser, deser;
// TreeNode* ans = deser.deserialize(ser.serialize(root));

LL(1)型文法 + 递归下降解析

思路：

我们可以构造二叉树序列化为 (leftTree)num(rightTree)，其中numroot的值。若子树为空，则用'#'表示。
那么该文法为

T -> (T) num (T)
	| #

其无左递归也不存在二义性，故为LL(1)型文法，那么我用LL(1)型文法的手工解析，递归下降法进行解析。

时间复杂度：$O(n)$

代码：

/**
 * 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:

    // Encodes a tree to a single string.
    string serialize(TreeNode* root) {
        if(root == NULL) return "#";
        string left = "(" + serialize(root->left) + ")";
        string right = "(" + serialize(root->right) + ")";
        return left + to_string(root->val) + right;
    }

    TreeNode *parse(string &data, int &cur) {
        if(data[cur] == '#') {
            cur ++;
            return NULL;
        }
        TreeNode *root = new TreeNode();
        root->left = parseSub(data, cur);
        root->val = parseInt(data, cur);
        root->right = parseSub(data, cur);
        return root;
    }

    TreeNode *parseSub(string &data, int &cur) {
        cur ++;
        TreeNode *root = parse(data, cur);
        cur ++;
        return root;
    }

    int parseInt(string &data, int &cur) {
        int x = 0, sign = 1;
        if(!isdigit(data[cur])) {
            cur ++;
            sign = -1;
        }
        while(isdigit(data[cur])) {
            x = x * 10 + (data[cur ++] - '0');
        }
        return x * sign;
    }

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

// Your Codec object will be instantiated and called as such:
// Codec ser, deser;
// TreeNode* ans = deser.deserialize(ser.serialize(root));