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606. Construct String from Binary Tree | leetcode solution

 606Construct String from Binary Tree

Given the root of a binary tree, construct a string consisting of parenthesis and integers from a binary tree with the preorder traversal way, and return it.

Omit all the empty parenthesis pairs that do not affect the one-to-one mapping relationship between the string and the original binary tree.

 

Example 1:

Input: root = [1,2,3,4]
Output: "1(2(4))(3)"
Explanation: Originally, 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: root = [1,2,3,null,4]
Output: "1(2()(4))(3)"
Explanation: Almost the same as the first example, except we cannot omit the first parenthesis pair to break the one-to-one mapping relationship between the input and the output.

 

Constraints:

  • The number of nodes in the tree is in the range [1, 104].
  • -1000 <= Node.val <= 1000

solution:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) 
 * };
 */
class Solution {
public:
    string tree2str(TreeNode* root) { 
        if(root==NULL)
            return "";
        string a = to_string(root->val);
        if(root->left==NULL && root->right==NULL)
            return a+"";
        if(root->right==NULL)
            return a+"("+tree2str(root->left)+")";
        return a+"("+tree2str(root->left)+")("+tree2str(root->right)+")"; 
    }
};



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