Leetcode606. Construct String from Binary Tree-优快云博客

本文链接：https://blog.youkuaiyun.com/mohuak/article/details/78427302

本文介绍LeetCode上的第606题Construct String from Binary Tree的解决方案，该题要求从二叉树中构建字符串，采用递归算法实现。通过两个示例详细解释了如何忽略不必要的空括号，确保输入输出的一一对应关系。

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Leetcode606. Construct String from Binary Tree

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.

题目分析：关于树的算法题，首先第一步考虑递归算法。这道题也不例外，也是用递归去实现。

代码：

class Solution {
public:
    string tree2str(TreeNode* t) {
        if (t == NULL) return "";
        string s = to_string(t->val);
        if (t->left) s += '(' + tree2str(t->left) + ')';
        else if (t->right) s += "()";
        if (t->right) s += '(' + tree2str(t->right) + ')';
        return s;
    }
};