剑指 Offer 37. 序列化二叉树

 

 

思路#

这里使用层序遍历产生的序列当成序列化，再使用此序列层序创建二叉树即对应反序列化。

使用“#”代表空节点，并且各节点的值在序列化的字符串中以空格分隔。

注：本题解法不唯一，如使用：先序遍历进行序列化 + 先序创建反序列化 也是可以的。

/**
 * 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.
    //层序遍历二叉树, #代表空节点
    string serialize(TreeNode* root) {
        if(root == NULL)
            return "";

        string res = "";
        queue<TreeNode*> Q;
        Q.push(root);
        while(!Q.empty()) {
            TreeNode *t = Q.front();
            Q.pop();
            
            if(t != NULL) {
                res += to_string(t->val) + " ";
                Q.push(t->left);
                Q.push(t->right);
            } else {
                res += "# ";
            }
        }
        return res;
    }

    // Decodes your encoded data to tree.
    //层序创建二叉树
    TreeNode* deserialize(string data) {
        if(data == "")
            return NULL;
        
        istringstream is(data);
        string value;
        is >> value;
        TreeNode *node = new TreeNode(stoi(value));
        queue<TreeNode*> Q;
        Q.push(node);
        while(!Q.empty()) {
            TreeNode *t = Q.front();
            Q.pop();
            
            string lvalue, rvalue;
            is >> lvalue;
            is >> rvalue;

            if(lvalue == "#") {
                t->left = NULL;
            } else {
                t->left = new TreeNode(stoi(lvalue));
                Q.push(t->left);
            }

            if(rvalue == "#") {
                t->right = NULL;
            } else {
                t->right = new TreeNode(stoi(rvalue));
                Q.push(t->right);
            }
        }

        return node;
    }
};

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