题解 | #序列化二叉树#

通过前序遍历或层次遍历（递归 或 非递归）去建树

注意char* 的使用

/*
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};
*/
class Solution {
public:
    char* Serialize(TreeNode *root) {    
        if(!root)
            return NULL;
        queue<TreeNode*> q;
        int* s = new int[101];
        int count = 0;
        q.push(root);
        while(!q.empty()){
            TreeNode* temp = q.front();
            if(temp)
                s[count] = temp->val;
            else
                s[count] = -1;
            count ++;
            if(temp){
                q.push(temp->left);
                q.push(temp->right);
            }
            q.pop();
        }
        return (char*)s;
    }
    TreeNode* Deserialize(char *str) {
        if(!str)
            return NULL;
        int* t = (int*)str;
        TreeNode* root = new TreeNode(t[0]);
        queue<TreeNode*> q;
        int count = 1;
        q.push(root);
        while(!q.empty()){
            TreeNode* temp = q.front();
            if(t[count] != -1){
                temp->left = new TreeNode(t[count]);
                q.push(temp->left);
            }
            count ++;
            if(t[count] != -1){
                temp->right = new TreeNode(t[count]);
                q.push(temp->right);
            }
            count ++;
            q.pop();
        }
        return root;
    }
};