PROBLEM

剑指 Offer 37. 序列化二叉树

难度 困难

MY ANSWER

  • 队列实现bfs，使用to_string来实现int到string的转换，从而实现二叉树的序列化。
  • 首先使用循环与substr对字符串进行分割，储存到string向量中。同样使用队列实现二叉树的构建，从队头中取节点添加儿子节点，将儿子节点入队。同时遍历string向量，index指示当前节点对应的儿子节点的值，构建节点使用stoi将字符串转换为int。
  • 时间复杂度O(n)，空间复杂度O(n)。
/**
 * 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) {
            return "[]";
        }
        string str = "[";
        queue<TreeNode*> bfs;
        bfs.push(root);
        int cnt = 0;
        while(bfs.size()) {
            cnt++;
            TreeNode* tmp = bfs.front();
            if(tmp) {
                str += to_string(tmp->val) + ",";
                bfs.push(tmp->left);
                bfs.push(tmp->right);
            }
            else {
                str += "null,";
            }
            bfs.pop();
        }
        str[str.length() - 1] = ']';
        //cout<<str<<endl;
        return str;
    }

    // Decodes your encoded data to tree.
    TreeNode* deserialize(string data) {
        TreeNode* root = NULL;
        if(!data.compare("[]")) {
            return root;
        }
        vector<string> seq;
        int start = 1, end = 1;
        for(int i = 1; i < data.length(); i++) {
            if(data[i] == ',' || data[i] == ']') {
                end = i;
                seq.push_back(data.substr(start, end - start));
                start = end + 1;
            }
        }
        // for(int i = 0; i < seq.size(); i++) {
        //     cout<<seq[i]<<endl;
        // }
        root = new TreeNode(stoi(seq[0]));
        queue<TreeNode*> bfs;
        bfs.push(root);
        int index = 1;
        while(index < seq.size() && bfs.size()) {
            TreeNode* tmp = bfs.front();
            if(!tmp) {
                bfs.pop();
                continue;
            }
            if(seq[index].compare("null")) {
                tmp->left = new TreeNode(stoi(seq[index]));
            }
            else {
                tmp->left = NULL;
            }
            bfs.push(tmp->left);
            index++;
            if(seq[index].compare("null")) {
                tmp->right = new TreeNode(stoi(seq[index]));
            }
            else {
                tmp->right = NULL;
            }
            bfs.push(tmp->right);
            index++;
            bfs.pop();
        }
        return root;
    }
};

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

BETTER SOLUTION

思路相同。

SUMMARY

bfs实现对树的层序遍历，注意to_string、stoi、substr这几个库函数的调用。