Problem

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.

Examples

Example 1:

		1
	  /   \
	 2     3
	/
   4

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:

		1
	  /   \
	 2     3
	  \
       4

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.

Solution

Method 1 - Inorder Traversal

Code

public String tree2str(TreeNode t) {
	if (t == null) {
		return "";
	}
	String left = tree2str(t.left);
	String right = tree2str(t.right);
	if (left.equals("") && right.equals("")) {
		return String.valueOf(t.val);
	}
	// we have to check empty for left side only, as right side can be skipped
	if (left.equals("")) {
		left = "()";
	} else {
		left = "(" + left + ")";
	}
	if (!right.equals("")) {
		right = "(" + right + ")";
	}
	return t.val + left + right;
}

Complexity

  • Time: O(N)
  • Space: O(N)