Need help constructing string with parenthesis from Binary Tree

Question

I am trying to solve the following algorithm question:

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 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: Originallay 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.

I have written the following code:

class TreeNode {
    constructor(val, left, right) {
        this.val = (val === undefined ? 0 : val)
        this.left = (left === undefined ? null : left)
        this.right = (right === undefined ? null : right)
    }
}

const tree2str = (t) => {
    const result = []
    const stack = []

    const dfs = (current) => {
        if (current === null) return
        result.push(current.val)
        if (!current.left && current.right) result.push('(')
        if (current.left) {
            result.push('(')
            stack.push(')')
            dfs(current.left)
        }
        while (stack.length) {
            result.push(stack.pop())
        }
        if (!current.left && current.right) stack.push(')')
        if (current.right) {
            result.push('(')
            stack.push(')')
            dfs(current.right)
        }
    }
    dfs(t)
    return result.join('')
}

The test cases I have so far:

const tree = new TreeNode(1, new TreeNode(2, new TreeNode(4)), new TreeNode(3))
const tree2 = new TreeNode(1, new TreeNode(2, null, new TreeNode(4)), new TreeNode(3))
const tree3 = new TreeNode(1, new TreeNode(2, new TreeNode(3), new TreeNode(4)))

console.log(tree2str(tree)) // "1(2(4)())(3()())" ==> "1(2(4))(3)"
console.log(tree2str(tree2)) // "1(2()(4))(3)"
console.log(tree2str(tree3)) // "1(2(3)(4))" instead got "1(2(3))(4)"

Only two work, however, I am having trouble with the third, and can't spot where I am going wrong.

Answer 1

The stack makes things a bit more complicated. The bug was that eventually you would pop the stack twice in your last test case and that would mess up the open-close parenthesis. I removed the stack and made the code a bit easier to debug. However, I have used the dual-stack approach for interpreting arithmetic expression, but that seems unnecessary here.

You can try the code out at this sandbox: https://codesandbox.io/s/need-help-constructing-string-with-parenthesis-from-binary-tree-p3sk3?file=/src/index.js

const tree2str = t => {
  const result = [];
  const dfs = current => {
    if (!current) {
      return;
    }

    const { val, left, right } = current;
    if (val !== null && val !== undefined) {
      result.push(val);
      if (left) {
        result.push("(");
        dfs(left);
        result.push(")");
      } else if (!left && right) {
        result.push("()");
      }

      if (right) {
        result.push("(");
        dfs(right);
        result.push(")");
      }
    }
  };
  dfs(t);
  return result.join("");
};

This version seems to produce the results you were looking for. That was fun though!