Problem #606 (Construct String from Binary Tree | Binary Tree, Depth-First Search, String, 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

Solutions

Breadth-First Search(Recursive)

Code

• Java

  class Solution {
    public String tree2str(TreeNode root) {
        if(root == null) return "";
        String s = String.valueOf(root.val);
        String left = tree2str(root.left);
        String right = tree2str(root.right);
        if(left == "" && right == "") return s;
        if(right == "") return s + "(" + left + ")";
        if(left == "") return s + "()" + "(" + right + ")";
        return s + "(" + left + ")" + "(" + right + ")";
    }
  }
  

  

Complexity

• Time O(log n)
• Space O(1)