[Swift]LeetCode331. 驗證二叉樹的前序序列化 | Verify Preorder Serialization of a Binary Tree

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公衆號：山青詠芝（shanqingyongzhi）
➤博客園地址：山青詠芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：http://www.javashuo.com/article/p-rzwsmbqi-hy.html 
➤若是連接不是山青詠芝的博客園地址，則多是爬取做者的文章。
➤原文已修改更新！強烈建議點擊原文地址閱讀！支持做者！支持原創！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

One way to serialize a binary tree is to use pre-order traversal. When we encounter a non-null node, we record the node's value. If it is a null node, we record using a sentinel value such as #.

     _9_
    /   \
   3     2
  / \   / \
 4   1  #  6
/ \ / \   / \
# # # #   # #

For example, the above binary tree can be serialized to the string "9,3,4,#,#,1,#,#,2,#,6,#,#", where # represents a null node.

Given a string of comma separated values, verify whether it is a correct preorder traversal serialization of a binary tree. Find an algorithm without reconstructing the tree.

Each comma separated value in the string must be either an integer or a character '#' representing null pointer.

You may assume that the input format is always valid, for example it could never contain two consecutive commas such as "1,,3".

Example 1:

Input: 
Output: "9,3,4,#,#,1,#,#,2,#,6,#,#"true

Example 2:

Input: 
Output: 
"1,#"false

Example 3:

Input: 
Output: "9,#,#,1"false

---

序列化二叉樹的一種方法是使用前序遍歷。當咱們遇到一個非空節點時，咱們能夠記錄下這個節點的值。若是它是一個空節點，咱們能夠使用一個標記值記錄，例如 #。

     _9_
    /   \
   3     2
  / \   / \
 4   1  #  6
/ \ / \   / \
# # # #   # #

例如，上面的二叉樹能夠被序列化爲字符串 "9,3,4,#,#,1,#,#,2,#,6,#,#"，其中 # 表明一個空節點。

給定一串以逗號分隔的序列，驗證它是不是正確的二叉樹的前序序列化。編寫一個在不重構樹的條件下的可行算法。

每一個以逗號分隔的字符或爲一個整數或爲一個表示 null 指針的 '#'。

你能夠認爲輸入格式老是有效的，例如它永遠不會包含兩個連續的逗號，好比 "1,,3" 。

示例 1:

輸入: 
輸出: "9,3,4,#,#,1,#,#,2,#,6,#,#"true

示例 2:

輸入: 
輸出: 
"1,#"false

示例 3:

輸入: 
輸出: "9,#,#,1"false

---

72ms

 1 class Solution {
 2     func isValidSerialization(_ preorder: String) -> Bool {
 3         let set = CharacterSet(charactersIn: ",")
 4         let preorderArr = preorder.components(separatedBy: set)
 5         
 6         if preorderArr.count % 2 == 0 {
 7             return false
 8         }
 9         
10         var i = 1
11         
12         for c in preorderArr {
13             if i == 0 {
14                 return false
15             }
16             if c == "#" {
17                  i -= 1
18             }else {
19                 i += 1
20             }
21         }
22         
23         return i == 0
24     }
25 }

---

76ms

 1 class Solution {
 2     func isValidSerialization(_ preorder: String) -> Bool {
 3         let arr = preorder.components(separatedBy:",")
 4         if arr[0] == "#" && arr.count != 1 {
 5             return false
 6         }
 7         var count = 0
 8         
 9         for i in 0..<arr.count {
10             let str = arr[i]
11             
12             if str == "#" {
13                 count += 1
14             } else {
15                 count -= 1
16             }
17             
18             if count == 1 && i != arr.count - 1 {
19                 return false
20             }
21         }
22         
23         return count == 1
24     }
25 }

---

80ms

 1 class Solution {
 2     func isValidSerialization(_ preorder: String) -> Bool {
 3  let nodes = preorder.components(separatedBy: ",")
 4         var stack = [String]()
 5         
 6         for i in 0..<nodes.count {
 7             let node = nodes[i]
 8             while node == "#" && !stack.isEmpty && stack.last! == "#" {
 9                 stack.removeLast()
10                 if stack.isEmpty {
11                     return false
12                 }
13                 stack.removeLast()
14             }
15             stack.append(node)
16         }
17         
18         return stack.count == 1 && stack.last! == "#"
19     }
20 }

---

140ms

 1 class Solution {
 2     func isValidSerialization(_ preorder: String) -> Bool {
 3         let res = preorder.components(separatedBy: ",")
 4         var diff = 1
 5         
 6         for r in res {
 7             diff -= 1
 8             if diff < 0 {
 9                 return false
10             }
11             if r != "#" {
12                 diff += 2
13             }
14         }
15         
16         return diff == 0
17     }
18 }