Application scenarios of Catalan number

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Carter LAN number

Derivation of Cartland number ：
Carter LAN number （Catalan）

Application scenarios

1. Suppose there is n A left bracket and n Right parenthesis , How many legal combinations do they have .
   Altogether C(2n,n) Combinations of , Suppose these combinations are A aggregate .
   For illegal combinations , There must be a prefix , The number of right parentheses is more than that of left parentheses 1 individual , such as ())(() Chinese prefix ()), And the suffix must be the right bracket less than the left bracket 1 One ((), We reverse the suffix （ The left bracket becomes the right bracket , The right bracket becomes the left bracket ）, In this way, the right bracket in the whole combination is n+1, The left bracket is n-1. There are C(2n,n-1) Combinations of , Suppose they are B aggregate .
   A Illegal combinations in can be pushed through inversion B All cases in the set , and B Collections can also be pushed out by reversing A Illegal combination in . therefore A The illegal quantity in is equal to C(2n,n-1).
   Sum up , There are legal combinations C(2n,n)-C(2n,n-1) Kind of .

2. n How many legal ways to get in and out of the stack .
   The number of any prefix out of the stack must be less than the number of prefix in the stack , So this problem turns into the bracket problem above . Its combination is also C(2n,n)-C(2n,n-1) Kind of .

3. Altogether n Nodes , How many ways to form a binary tree .
   Yes 0 Nodes , The method is empty tree , The way to form a binary tree 1 Kind of .
   Yes 1 Nodes , The tree is for itself ,1 Kind of .
   Yes 2 Nodes , root + Left or root + Right ,2 Kind of .
   Yes n Nodes , Select a node as the head node ; The left side of the head node 0 Nodes , On the right n-1 Nodes ; The left side of the head node 1 Nodes , On the right n-2 Nodes ; The left side of the head node 2 Nodes , On the right n-3 Nodes ... Can be launched k(n)=k(0)*k(n-1)+k(1)*k(n-2)+....+k(n-2)*k(1)+k(n-1)*k(0), That is to say, the number of Cartland C(2n,n)-C(2n,n-1) Combination mode .