Leetcode 606 - Construct String from Binary Tree

Note:

  • Given traveral is in level order traversal.
  • Normally it’d be quite easy, but when do we need to omit () and when not?
  • Ask ourself when do we need null in our input?
    • Ans: when we need to tell machine that left subtree is empty and the next node is for right subtree.
  • So in our recursion, when a node has no left subtree but has right subtree, we cannot omit ().

Question:

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:

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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)"

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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.

Code:

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/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {string}
*/
var tree2str = function(root) {
const ans = dfs(root);
return ans.slice(1, ans.length - 1);

function dfs(node) {
if (!node) return '';
const left = dfs(node.left);
const right = dfs(node.right);
// Check if `left` is null.
return '(' + node.val + (node.right && !node.left ? '()' : left) + right + ')';
}
};