606. Construct String from Binary Tree 二叉树的字符串描述

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#二叉树 #string

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本文介绍了一种将二叉树转换为字符串的算法，遵循先序遍历的原则，并通过递归实现。对于空节点使用空括号对表示，但会省略不影响一到一映射关系的多余空括号。提供了具体的示例来展示如何从二叉树结构生成相应的字符串。

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.
根据给定的二叉树返回根据其建立的字符串，建立原则按照先序序列顺序，有子节点的用括号括起来，叶子节点的括号省略。特别的，若左孩子不存在而有孩子存在，则左孩子的括号不可省略。
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
递归调用即可。

/**
 * 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==nullptr) return "";
        string res;
        res+=to_string(t->val);
        if(t->left) res+="("+tree2str(t->left)+")";
        else if( t->right) res+="()";
        if(t->right) res+="("+tree2str(t->right)+")";
        return res;
    }
};