LeetCode 606. Construct String from Binary Tree根据二叉树创建字符串 (C++)

题目：

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.

分析：

给定一颗二叉树，按要求创建字符串。

我们可以发现，一个节点如果有左右节点的话，是要用()括起来左右节点，如果left存在，righ不存在，可以省略掉右边的括号，如果left不存在，right存在，左括号是不能省略的。

程序：

class Solution {
public:
    string tree2str(TreeNode* t) {
        string s = "";
        if(t == nullptr) return s;
        if(!t->left && !t->right)
            return s + to_string(t->val);
        if(t->right == nullptr)
            return s + to_string(t->val) + "(" + tree2str(t->left) + ")";
        return s + to_string(t->val) + "(" + tree2str(t->left) + ")("+ tree2str(t->right) + ")";
    }
};

LeetCode 606. Construct String from Binary Tree根据二叉树创建字符串 (C++)

原文：https://www.cnblogs.com/silentteller/p/10747874.html