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 * } */functree2str(root *TreeNode) string {if root ==nil {return"" }if root.Left ==nil&& root.Right ==nil {return strconv.Itoa(root.Val) } elseif root.Left ==nil {return strconv.Itoa(root.Val) +"()"+"("+ tree2str(root.Right) +")" } elseif root.Right ==nil{return strconv.Itoa(root.Val) +"("+ tree2str(root.Left) +")" }return strconv.Itoa(root.Val) +"("+ tree2str(root.Left) +")"+"("+ tree2str(root.Right) +")"}