题解 | #序列化二叉树#

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 * };
 */
/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 *
 * @param root TreeNode类 
 * @return char字符型一维数组
 */
char* Serialize(struct TreeNode* root ) {
    // write code here
    //层序
    if(!root) return NULL;
    char* str = (char*)malloc(sizeof(char)*300);
    struct TreeNode** pr = (struct TreeNode**)malloc(sizeof(struct TreeNode*)*100);
    int p1 = -1 , p2 = -1 , pa = -1;
    pr[++p1] = root;
    while(p1 != p2){
        root = pr[++p2];
        if(root){
           str[++pa] =root->val;
           pr[++p1] = root->left;
           pr[++p1] = root->right;
        }
        else{
            str[++pa] = '#';
        }
    }
    str[++pa] = '\0';
    free(pr);
    return str;
}
/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * 
 * @param str char字符型一维数组 
 * @return TreeNode类
 */
struct TreeNode* Deserialize(char* str ) {
    // write code here
    if(!str) return NULL;
    struct TreeNode* root = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    struct TreeNode* pro[100];
    int p1 = -1 , p2 = -1;
    root->val = (int)*str;
    str++;
    pro[++p1] = root;
    while(*str != '\0'){
        struct TreeNode* t = pro[++p2];
        if(*str != '#'){
            struct TreeNode* r = (struct TreeNode*)malloc(sizeof(struct TreeNode));
            r->val = (unsigned char)*str;
            pro[++p1] = r;
            t->left = r;
        }
        else{
            t->left = NULL;
        }
        str++;
        if(*str != '#'){
            struct TreeNode* r = (struct TreeNode*)malloc(sizeof(struct TreeNode));
            r->val = (unsigned char)*str;
            pro[++p1] = r;
            t->right = r;
        }
        else{
            t->right = NULL;
        }
        str++;
    }
    return root;
}