剑指OFFER----面试题37. 序列化二叉树

链接:https://leetcode-cn.com/problems/xu-lie-hua-er-cha-shu-lcof/

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:

    string serialize(TreeNode* root) {
        string res;
        dfs_s(root, res);
        return res;
    }

    void dfs_s(TreeNode *root, string &res)
    {
        if (!root) {
            res += "null ";
            return;
        }
        res += to_string(root->val) + ' ';
        dfs_s(root->left, res);
        dfs_s(root->right, res);
    }

    TreeNode* deserialize(string data) {
        int u = 0;
        return dfs_d(data, u);
    }

    TreeNode* dfs_d(string data, int &u)
    {
        if (u == data.size()) return NULL;
        int k = u;
        while (data[k] != ' ') k ++ ;
        if (data[u] == 'n') {
            u = k + 1;
            return NULL;
        }
        int val = 0, sign = 1;
        if (u < k && data[u] == '-') sign = -1, u ++ ;
        for (int i = u; i < k; i ++ ) val = val * 10 + data[i] - '0';
        val *= sign;
        u = k + 1;
        auto root = new TreeNode(val);
        root->left = dfs_d(data, u);
        root->right = dfs_d(data, u);
        return root;
    }


};

// Your Codec object will be instantiated and called as such:
// Codec codec;
// codec.deserialize(codec.serialize(root));
原文地址:https://www.cnblogs.com/clown9804/p/12389958.html