第十六周：( LeetCode606) Construct String from Binary Tree（c++）

原题：

You need to construct a string consists of parenthesis and integers from a binary tree with the preorder traversing way.

The null node needs to be represented by empty parenthesis pair “()”. And you need to omit all the empty parenthesis pairs that don’t affect the one-to-one mapping relationship between the string and the original binary tree.

Example 1:Input: Binary tree: [1,2,3,4]       1     /   \    2     3   /      4     Output: "1(2(4))(3)"Explanation: Originallay it needs to be "1(2(4)())(3()())", but you need to omit all the unnecessary empty parenthesis pairs. And it will be "1(2(4))(3)".

Example 2:Input: Binary tree: [1,2,3,null,4]       1     /   \    2     3     \        4 Output: "1(2()(4))(3)"Explanation: Almost the same as the first example, except we can't omit the first parenthesis pair to break the one-to-one mapping relationship between the input and the output.

思路：
本题的本质就是树的先序遍历，因为要加上括号所以要注意一些细节的处理，也涉及到一些c++字符串的操作，要注意区别好例子一和例子二。

代码：

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    string construct(TreeNode* t){        string s;        s+=("("+to_string(t->val));        if((t->left==NULL)&&(t->right==NULL))            s+=(")");        else if((t->left!=NULL)&&(t->right==NULL))            s+=(construct(t->left)+")");        else if((t->left==NULL)&&(t->right!=NULL)){            s+="()";            s+=(construct(t->right)+")");        }else            s+=(construct(t->left)+construct(t->right)+")");        return s;    }    string tree2str(TreeNode* t) {        string s;        if(t==NULL) return s;        //去除第一位“（”和最后一位“）”        return construct(t).substr(1,construct(t).length()-2);    }};