本文主要是介绍191111题，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
#include<iostream>
using namespace std;
struct TreeNode {int value;TreeNode*left;TreeNode*right;TreeNode(int x = 0) :value(x), right(NULL), left(NULL) {}
};
class Solution {
public:void Serialize(TreeNode*proot)//前序为例{if (proot == NULL){cout << "$ ";return;}//交换以下三句的顺序就可以可以得到前序、中序、后序序列cout << proot->value << " ";Serialize(proot->left);Serialize(proot->right);}TreeNode* DeSerialize(char **str)//递归时改变了str值使其指向后面的序列，因此声明char**{if (str == NULL)return NULL;TreeNode*root = DeSerializeCore(str);return root;}
private://由前序遍历序列反序列化为例TreeNode* DeSerializeCore(char **str){if (**str == '$'){(*str)++;return NULL;}int num = 0;if (**str != '\0')num = ((**str) - '0');TreeNode*root = new TreeNode(num);(*str)++;if (**str == '\0')return root;root->left = DeSerializeCore(str);root->right = DeSerializeCore(str);//递归结束后返回大树的根return root;}
};int main() {TreeNode*root = new TreeNode(1);TreeNode*node1 = new TreeNode(2); TreeNode*node2 = new TreeNode(3);TreeNode*node3 = new TreeNode(4); TreeNode*node4 = new TreeNode(5); TreeNode*node5 = new TreeNode(6);root->left = node1; root->right = node2;node1->left = node3;node2->left = node4; node2->right = node5;Solution solution;solution.Serialize(root);cout << endl;char*str = "124$$$35$$6$$";TreeNode*reversed_root = solution.DeSerialize(&str);cout << reversed_root->right->left->value << endl;system("pause");return 0;
}
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