[Leetcode] 606. Construct String from Binary Tree 解题报告

二叉树前序遍历字符串化

最新推荐文章于 2022-08-11 10:15:00 发布

原创最新推荐文章于 2022-08-11 10:15:00 发布 · 275 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#Leetcode #解题报告 #Tree

IT公司面试习题专栏收录该内容

736 篇文章

订阅专栏

本文介绍了一种将二叉树通过前序遍历转换为特定字符串的方法，该字符串由括号和整数组成，同时探讨了如何省略不影响一对一映射关系的空括号对。

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.

思路：

又一道练手题目，哈哈。祝大家面试的时候都能遇到这种题目！

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 ret(to_string(t->val));
        if (!t->left && !t->right) {    // t is a leaf, so do not add parenthesis at all
            return ret;
        }
        ret += '(';                     // no matter t has a left child or not, we have to add parenthesis
        ret += tree2str(t->left);
        ret += ')';
        if (t->right) {                 // we add parenthesis only when t has a right child
            ret += '(';
            ret += tree2str(t->right);
            ret += ')';
        }
        return ret;
    }
};