lintcode binary-tree-serialization 二叉树的序列化与反序列化

问题描述

lintcode

参考

Serialization/Deserialization of a Binary Tree
LintCode Binary Tree Serialization

笔记

可以用先序遍历 和 使用"#"来表示NULL的方法来序列化，最后也用先序遍历来反序列化。

代码

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */
class Solution {
public:
    /**
     * This method will be invoked first, you should design your own algorithm 
     * to serialize a binary tree which denote by a root node to a string which
     * can be easily deserialized by your own "deserialize" method later.
     */
    string serialize(TreeNode *root) {
        // write your code here
        string s = "";
        writeTree(s, root);
        return s;
    }

    void writeTree(string &s, TreeNode* root)
    {
        if (root == NULL)
        {
            s += "# ";
            return;
        }
        s += (to_string(root->val) + ' ');
        writeTree(s, root->left);
        writeTree(s, root->right);
    }

    /**
     * This method will be invoked second, the argument data is what exactly
     * you serialized at method "serialize", that means the data is not given by
     * system, it's given by your own serialize method. So the format of data is
     * designed by yourself, and deserialize it here as you serialize it in 
     * "serialize" method.
     */
    TreeNode *deserialize(string data) {
        // write your code here
        int pos = 0;
        return readTree(data, pos);

    }

    TreeNode* readTree(string data, int& pos)
    {
        if (data[pos] == '#')
        {
            pos += 2;
            return NULL;
        }
        int nownum = 0;
        while (data[pos] != ' ')
        {
            nownum = nownum * 10 + (data[pos] - '0');
            pos++;
        }
        pos++;
        TreeNode* nowNode = new TreeNode(nownum);
        nowNode->left = readTree(data, pos);
        nowNode->right = readTree(data, pos);
        return nowNode;
    }
};