【LeetCode 606】 Construct String from Binary Tree (C++)

最新推荐文章于 2021-10-31 16:42:17 发布

原创最新推荐文章于 2021-10-31 16:42:17 发布 · 195 阅读

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#LeetCode #二叉树 #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，并使用特定规则来省略不必要的空节点括号，确保字符串与原始二叉树具有一一对应的关系。

Problem：

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

Solution:

/**
 * 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 s=to_string(t->val);
        if(t->left!=NULL){
            s+="("+tree2str(t->left)+")";
            if(t->right!=NULL)s+="("+tree2str(t->right)+")";
        }else if(t->right!=NULL)
            s+="()("+tree2str(t->right)+")";
        return s;
    }
};