二叉樹公共祖先節點的查找

  

#include <iostream>
#include <vector>
using std::endl;
using std::vector;
using std::cin;
using std::cout;
struct treeNode
{
int value;
treeNode*left;
treeNode*right;
};
<strong><span style="font-size:18px;">bool isAncestor(treeNode*root,treeNode*p)
{
	if(root==NULL)
	return false;
	if(root==p)
	  return true;
	return isAncestor(root->left,p)||isAncestor(root->right,p);
}
treeNode* findCommonAncestor(treeNode* root,treeNode* p,treeNode *q)
{
	 if(root==NULL)
	 return NULL;
	 if(root==p||root==q)
	    return root;
	bool is_p_left=isAncestor(root->left,p);//p在左子樹 
    bool is_q_left=isAncestor(root->left,q);//q在左子樹  
	if (is_p_left!=is_q_left)//p、q異側  
      return root;
    return is_p_left?findCommonAncestor(root->left,p,q):findCommonAncestor(root->right,p,q);
} 
treeNode *findAncestor(treeNode* root,treeNode*p, treeNode*q)
{
	if(!isAncestor(root,p)&&!isAncestor(root,q))
        return NULL;
      return  findCommonAncestor(root,p,q);
 }</span></strong>
 treeNode * treeInsert(treeNode* head,int n)  
{    
    treeNode* newNode=new treeNode;  
    newNode->left=newNode->right=NULL; 
    newNode->value=n;  
    if (head==NULL)  
    {    
          
          return newNode;  
    }  
    treeNode *head1=head,*head2=NULL;  
        while (head1!=NULL)//先確定待插入的父親節點   
             {     
               head2=head1;  
                if(head1->value>n)  
                    head1=head1->left;  
                else  
                   head1=head1->right;  
                  
             }   
             if(head2->value>n)  
                {  
                    head2->left=newNode;  
                      
                }  
                  
                else   
                {  
                     head2->right=newNode;  
                }  
                   
          return head;    
} 
int getHeight(treeNode * root)
 {
     if(root==NULL)
	     return 0;    
  else 
   return (getHeight(root->left)>getHeight(root->right)?getHeight(root->left):getHeight(root->right))+1;
} 
void  ByLevel(treeNode * root,int level)
{
  if(root==NULL||level<0)
       return ;
   if (level==0)
       cout<<root->value<<" ";
   else 
    {
       ByLevel(root->left,level-1);
	   ByLevel(root->right,level-1);
	
    }    
	

}

int main()
{  
  treeNode *head=NULL;
	head=treeInsert(head,20);
	head=treeInsert(head,15);
	head=treeInsert(head,25);
	head=treeInsert(head,30);
	head=treeInsert(head,10);
	head=treeInsert(head,18);
	head=treeInsert(head,22);
	head=treeInsert(head,23);
	head=treeInsert(head,21);
	treeNode* s1=head->right->left->left;//
	  treeNode* s2=head->right->left->right;//30
	  int high=getHeight(head);
	  for(int i=0;i<high;i++)
	    {
	      cout<<"LEVEL:"<<i+1<<endl; 
		  ByLevel(head,i);
		  cout<<endl;
	      cout<<"__________________"<<endl; 
	    } 
   treeNode* s3=findAncestor(head,s1,s2);
        if(s3)
{   cout<<s2->value<<" and "<<s1->value<<" 's common ancestor is "; 
	cout<<s3->value;
}
    
	  return 0; 
}
/*
  時間複雜度分析：O（n）
   設有2*n個節點
   左子樹n個
   右子樹n個
   則isAncestor 第一次要遍歷2*n
    第二次2*n/2
	第三次2*n/2/2
	每次減半 
	因此累加爲O（n） 
*/