[Swift]LeetCode606. 根據二叉樹建立字符串 | Construct String from Binary Tree

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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.

---

你須要採用前序遍歷的方式，將一個二叉樹轉換成一個由括號和整數組成的字符串。

空節點則用一對空括號 "()" 表示。並且你須要省略全部不影響字符串與原始二叉樹之間的一對一映射關係的空括號對。

示例 1:

輸入: 二叉樹: [1,2,3,4]
       1
     /   \
    2     3
   /    
  4     

輸出: "1(2(4))(3)"

解釋: 本來將是「1(2(4)())(3())」，
在你省略全部沒必要要的空括號對以後，
它將是「1(2(4))(3)」。

示例 2:

輸入: 二叉樹: [1,2,3,null,4]
       1
     /   \
    2     3
     \  
      4 

輸出: "1(2()(4))(3)"
解釋: 和第一個示例類似，
除了咱們不能省略第一個對括號來中斷輸入和輸出之間的一對一映射關係。

---

Runtime: 80 ms

Memory Usage: 20.7 MB

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     public var val: Int
 5  *     public var left: TreeNode?
 6  *     public var right: TreeNode?
 7  *     public init(_ val: Int) {
 8  *         self.val = val
 9  *         self.left = nil
10  *         self.right = nil
11  *     }
12  * }
13  */
14 class Solution {
15     func tree2str(_ t: TreeNode?) -> String {
16         if t == nil {return String()}
17         var res:String = String(t!.val)
18         if t!.left == nil && t!.right == nil {return res}
19         res += "(" + tree2str(t!.left) + ")"
20         if t!.right != nil
21         {
22             res += "(" + tree2str(t!.right) + ")"
23         }
24         return res
25     }
26 }

---

84ms

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     public var val: Int
 5  *     public var left: TreeNode?
 6  *     public var right: TreeNode?
 7  *     public init(_ val: Int) {
 8  *         self.val = val
 9  *         self.left = nil
10  *         self.right = nil
11  *     }
12  * }
13  */
14 class Solution {
15     func tree2str(_ t: TreeNode?) -> String {
16         guard t != nil else {
17             return ""
18         }
19         var res = ""
20         inorderMakeString(t!, &res)
21         return res
22     }
23     
24     private func inorderMakeString(_ t: TreeNode?, _ str: inout String) {
25         
26         guard let node = t else {
27             return
28         }
29         str += String(node.val)
30         if node.left == nil && node.right == nil {
31             return
32         }
33 
34         if node.right != nil || node.left != nil {
35             str += "("
36             inorderMakeString(node.left, &str)
37             str += ")"
38 
39         }
40         if node.right != nil {
41             str += "("
42             inorderMakeString(node.right!, &str)
43             str += ")"
44         }
45     }
46 }

---

88ms

 1 class Solution {
 2     func tree2str(_ t: TreeNode?) -> String {
 3         guard let t = t else {
 4             return ""
 5         }
 6         if isLeaf(t) {
 7             let s = "\(t.val)"
 8             // print(s)
 9             return s
10         } else if let left = t.left, let right = t.right {
11             let s = "\(t.val)(\(tree2str(left)))(\(tree2str(right)))"
12             // print(s)
13             return s
14         } else if let left = t.left {
15             let s =  "\(t.val)(\(tree2str(left)))"
16             // print(s)
17             return s
18         } else if let right = t.right {
19             let s =  "\(t.val)()(\(tree2str(right)))"
20             // print(s)
21             return s
22         }
23         else {
24             return ""
25         }
26     }
27 
28     func isLeaf(_ t: TreeNode) -> Bool {
29         return t.left == nil && t.right == nil
30     }
31 }

---

92ms

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     public var val: Int
 5  *     public var left: TreeNode?
 6  *     public var right: TreeNode?
 7  *     public init(_ val: Int) {
 8  *         self.val = val
 9  *         self.left = nil
10  *         self.right = nil
11  *     }
12  * }
13  */
14 class Solution {
15     func tree2str(_ t: TreeNode?) -> String {
16         guard let node = t else {
17             return ""
18         }
19         var result = "\(node.val)"
20         if let left = node.left {
21             result+="(\(tree2str(left)))"
22         }
23         if let right = node.right {
24             if node.left == nil {
25                 result += "()"
26             }
27             result+="(\(tree2str(right)))"
28         }
29         return result
30     }
31 }

---

96ms

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     public var val: Int
 5  *     public var left: TreeNode?
 6  *     public var right: TreeNode?
 7  *     public init(_ val: Int) {
 8  *         self.val = val
 9  *         self.left = nil
10  *         self.right = nil
11  *     }
12  * }
13  */
14 class Solution {
15     func tree2str(_ t: TreeNode?) -> String {
16         guard let t = t else { return "" }
17         if t.right == nil && t.left == nil { return "\(t.val)" }
18         var toReturn = "\(t.val)(\(tree2str(t.left)))"
19         if t.right != nil {
20             toReturn += "(\(tree2str(t.right)))"
21         }
22         return toReturn
23     }
24 }

---

128ms

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     public var val: Int
 5  *     public var left: TreeNode?
 6  *     public var right: TreeNode?
 7  *     public init(_ val: Int) {
 8  *         self.val = val
 9  *         self.left = nil
10  *         self.right = nil
11  *     }
12  * }
13  */
14 class Solution {
15     func tree2str(_ t: TreeNode?) -> String {
16         var result: String = ""
17         preorder(node: t, result: &result)
18         return result
19     }
20     
21     private func preorder(node: TreeNode?, result: inout String) {
22         if node == nil { return }
23         result = result.appending(String(node!.val))
24         
25          if node?.left != nil || node?.right != nil {
26             result = result.appending("(")
27             preorder(node: node?.left, result: &result)
28             result = result.appending(")")
29         }
30         
31         if node?.right != nil {
32             result = result.appending("(")
33             preorder(node: node?.right, result: &result)
34             result = result.appending(")")
35         }
36     }
37 }