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Construct String from Binary Tree - MiniTV

Construct String from Binary Tree - मिनी टीवी

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 an 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: 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: 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.

 

Solution:

using System;
using System.Collections.Generic;
using System.Text;

namespace LeetCode.AskGif.Easy.String
{  
     public class TreeNode {
         public int val;
         public TreeNode left;
         public TreeNode right;
         public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
             this.val = val;
             this.left = left;
             this.right = right;
         }
     }
 
    public class Tree2strSoln
    {
        public string Tree2str(TreeNode t)
        {
            if (t == null) return "";
            var str = new StringBuilder();
            str.Append(t.val);
            if (t.left != null)
            {
                str.Append("(");
                str.Append(Tree2str(t.left));
                str.Append(")");
            }
            if(t.right != null)
            {
                if(t.left == null)
                {
                    str.Append("()");                    
                }

                str.Append("(");
                str.Append(Tree2str(t.right));
                str.Append(")");
            }
            return str.ToString();
        }
    }
}

 

Time Complexity: O(n)

Space Complexity: O(n)