Lintcode---二叉树的序列化和反序列化

/**
 * 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.
     */
     
    /*
    思路：感觉上用BFS更容易解决,在这里使用先根遍历来实现；
    参考：http://blog.csdn.net/waltonhuang/article/details/51979479
    本题目难点在于，里面穿插关于字符串和整数间的互相转换。
    在序列化时，空节点的表示，不同节点值之间的分割。
    在反序列化时，字符串每个字符遍历时的控制条件和操作，以及将字符串转换为整数；
    */
    
    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] != ‘ ‘){
            //这里‘ ‘是为了分离不同的数字；
            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;
    }
};