Construct String from Binary Tree (Leetcode 606)

Leetcode 606

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
  • 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

const tree2str = (root) => {
  // if no value just return
  if (!root) {
    return "";
  }

  // continue to traverse the tree
  const left = tree2str(root.left);
  const right = tree2str(root.right);

  // only output left if there is left or right val
  const leftOutput = left || right ? `(${left})` : "";

  // only output right if there is a right val
  const rightOutput = right ? `(${right})` : "";

  return `${root.val}${leftOutput}${rightOutput}`;
};

// [1,2,3,4];
const example1 = {
  val: "1",
  left: { val: "2", left: { val: "4" } },
  right: { val: "3" }
};

//[1,2,3,null,4]
const example2 = {
  val: "1",
  left: { val: "2", right: { val: "4" } },
  right: { val: "3" }
};

console.log("example1", tree2str(example1));
console.log("example2", tree2str(example2));

Big O -> O(n)

Explanation: we are traversing over every item in the tree (n)