606. Create a string based on a binary tree

You need to convert a binary tree into a string of parentheses and integers in the way of a pre-order traversal.

Empty nodes are denoted by a pair of empty parentheses "()". And you need to omit all empty brace pairs that do not affect the one-to-one mapping between the string and the original binary tree.

Example 1:

Input: Two fork tree: [1,2,3,4]
       1
     /   \
    2     3
   /    
  4     

output: "1 (2 (4)) (3)"

Interpretation: "1 (2 (4) ()) (3 ())",
after you omit all unnecessary empty brackets,
it will be "1 (2 (4)) (3)".

Example 2:

Input: Two fork tree: [1,2,3,null,4]
       1
     /   \
    2     3
     \  
      4 

output: "1 (2 () (4)) (3)"

Explanation: Similar to the first example ,
except that we cannot omit the first pair of parentheses to break the one-to-one mapping between input and output.

/** * 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 ans;
        void Preordertraversal (treenode* root)//pre-order traversal {int left = 0;
            if (root) {ans + = to_string (root->val);
                if (root->left! = NULL) {left = 1;//flag bit, whether there is an ans + = (';
                Preordertraversal (Root->left);
            Ans + = ') ';
                    if (root->right! = NULL) {if (left = = 0)//If there is no right-hand node directly to the node, the description should be output one ()
                Ans + = "()";
                Ans + = ' (';
                Preordertraversal (Root->right);
            Ans + = ') ';
        }}} string Tree2str (Treenode* t) {preordertraversal (t);
    return ans; }

};