331. Verify preorder serialization of a Binary Tree

Method One:

My own method, build a stack, each hit the number on the stack, every encounter #, will pop up a number, the last stack must be left a #.

1      Public Booleanisvalidserialization (String preorder) {2         if(Preorder = =NULL|| Preorder.length () = = 0) {3             return false;4         }5string[] res = Preorder.split (",");6stack<string> stack =NewStack<string>();7Stack.push (res[res.length-1]);8          for(inti = 0; i < res.length-1; i++) {9String each =Res[i];Ten             if("#". Equals (each)) { One                 if(!Stack.isempty ()) { A Stack.pop (); -}Else { -                     return false; the                 } -}Else { - Stack.push (each); -             } +         } -         return!stack.isempty () && Stack.peek (). Equals ("#"); +}

Time complexity is O (n)

2. Other people's methods

Each number can have two sides issued, each # will occupy an edge, each number itself will occupy an edge, but will produce two edges, so maintain a count, once less than 0 returns false.

Note that the following are:

1) This number initialization is 1, equivalent to a root node own 1 is 1, and produce two, so add up is 1;

2) Every time you encounter a number, you have to separate-1 and +2, if 1 is less than 0, return false

1      Public Booleanisvalidserialization (String preorder) {2string[] res = Preorder.split (",");3         intCount = 1;4          for(String each:res) {5             if(--count < 0) {6                 return false;7             }8             if(!" #". Equals (each)) {9Count + = 2;Ten             } One         } A         returnCount = = 0; -}

Time complexity is O (n)

Method Three:

Very good looking method, each leaf node structure is "number, #,#", so we put all of this format is replaced by a "#", constantly shrinking until the end should be a #.

1      Public Boolean isvalidserialization (String preorder) {2         String after = Preorder.replaceall ("\\d+,#,#", "#"); 3         return after.equals ("#") | | !after.equals (preorder) && isvalidserialization (after); 4     }

The third line, the reason is "!after.equals (preorder)", because such as an illegal tree "three-way", and there is no "number, #,#" Such a structure, so replaceall will not change it

Time complexity is O (n^2)

331. Verify preorder serialization of a Binary Tree