LeetCode 606. Construct String from Binary Tree（C++版）

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#先序遍历

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本文介绍了一种将二叉树转换为字符串的算法，该算法遵循先序遍历的原则，并为每个节点添加括号来表示其子树。文章通过两个实例详细解释了算法的工作原理及特殊情况处理。

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题目描述：

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.

题目分析：

本质上是树的先序遍历，但是额外给每个孩子加上了一对括号。需要注意的点是当只有左孩子时，右孩子处不需要加括号，但只有右孩子时，左孩子需要加括号。

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 tree2str(TreeNode* t) {
        if(t == NULL) return "";

        string res;
        res = to_string(t->val);  //int转为string

        if(t -> left != NULL || t-> right != NULL) { //如果左孩子为空，但是右孩子不为空，左孩子处需要加一对括号
            res += "(" + tree2str(t -> left) + ")";
        }
        if(t -> right != NULL) {
            res += "(" + tree2str(t -> right) + ")";
        }

        return res;
    }
};