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

本文介绍了一种从二叉树通过先序遍历方式构建包含括号和整数字符串的方法。空节点用()表示，并省略不影响字符串与原始二叉树一对一映射的多余空括号对。举例说明了不同输入情况下的输出结果。

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606. Construct String from Binary Tree

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 output.

Implementation

/**
 * 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) {
        string res;
        constructString(t, res);
        return res;
    }

    void constructString(TreeNode* root, string& res) {
        if(!root)
            return;    

        res.append(to_string(root->val));

        if(!root->left && !root->right)
            return;

        else if(!root->left) {
            res.append("()");
            res.append("(");
            constructString(root->right, res);
            res.append(")");
        }    
        else if(!root->right) {
            res.append("(");
            constructString(root->left, res);
            res.append(")");
        }
        else {
            res.append("(");
            constructString(root->left, res); 
            res.append(")");
            res.append("(");
            constructString(root->right, res);
            res.append(")");
        }
    }
};