606. Construct String from Binary Tree

Easy

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

解題

Runtime: 12 ms, faster than 81.67%

Memory Usage: 8.2 MB, less than 43.33%

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func tree2str(root *TreeNode) string {
    if root == nil {
        return ""
    }

    if root.Left == nil && root.Right == nil {
        return strconv.Itoa(root.Val)
    } else if root.Left == nil {
        return strconv.Itoa(root.Val) + "()" + "(" + tree2str(root.Right) + ")"
    } else if root.Right == nil{
        return strconv.Itoa(root.Val) +  "(" + tree2str(root.Left) + ")"
    }

    return strconv.Itoa(root.Val) + "(" + tree2str(root.Left) + ")" + "(" + tree2str(root.Right) + ")"
}