Determine whether sequential traversal of ordered binary trees is correct (summary of recursive calculation)

Is

#include <iostream>using namespace The nature of the binary tree of std;//*/*. Termination condition: 1. Start >= End returns True *. 2.S >= e Because there is no problem, it will be able to reach the s>=e situation.     Knowing the arrival of S >= E is true *.  1. The last node is root 2. The node that precedes root is its right subtree, which is more than root, and the left subtree is 3. Recursive call */bool Treehelper (int a[], int s, int e) {if (A = = NULL) return 0;if (S >= e) return 1;  In the absence of a right subtree, it may appear that s >= e int i = E-1;while (i >= s && a[e] <= a[i]) i--;if (!treehelper (a,i+1,e-1)) {return 0;} int k = I;while (i >= s && a[e] > A[i]) i--;if (S < i+1) return 0;//If the left dial hand tree is judged to be greater than the root value, it is not a sequential r of a sort binary tree. Eturn Treehelper (A,s, k);} int postorderresult (int a[], int n) {return treehelper (a,0,n-1);} int main () {int a[] = {1,5,7,6,9,11,10,8};int i = Postorderresult (a,sizeof (a)/sizeof (a[0])); cout <<i << std:: Endl;return 0;} As for the pre-order of the more simple traversal on the line. Middle order words bool Treehelper2 (int a[], int s, int e) {if (a = = NULL) return null;if (S >= e) return 1;int i = s+1;while (i <= E && A[s] > A[i]) i++;if (!treehelper2 (a,s+1,i-1)) {return 0;} INT k = i;while (i <= e && a[s] <= a[i]) i++;if (E > I-1) return 0;return treehelper2 (a,k,e);  }

The recursion here is very classic, I summed it up, but the summary of the point ...

1. Sub-problem

2. Termination conditions (including some borders)

Sub-problem: Each post-judgment is divided into its left sub-tree of the next right sub-tree of the post-sequential, and then left to the sub-tree can be handed down. This will find a sub-problem.

Termination conditions: When s>=e, there will be s>=e only the recursive to each child node can occur, when recursive to this, this post-order represents that there is no problem.

General form of Summary:

void func (int a, int b)//The parameters here are parameters that control the size of the sub-problem

{

if (...) return; //recursive termination conditions

The following is to solve the current problem, generally involves sub-issues

int i = func (a+x,b-x); This is the division of the sub-problem, but also the control sub-problem

Int J = func (a-x,a+x); //

return i*j; //finally returns the result of this whole problem based on the value returned by the child problem.

}

Entertain yourself, light spray!!

Determine whether sequential traversal of ordered binary trees is correct (summary of recursive calculation)