如何判断一棵二叉树树是否为另一棵二叉树的子树

#include <iostream>
#include<queue>
using namespace std;
struct treeNode
{
  int value;
  treeNode* left;
  treeNode* right;
  treeNode* parent;
};//节点结构
void  insert(treeNode **root,int value)
 {
 	treeNode* toBe=new treeNode;
   	toBe->value=value;
	 toBe->parent=NULL;
 	toBe->left=toBe->right=NULL;
 	if(*root==NULL)
 	 {
 	 	 *root=toBe;
 	      return ;
	  }
	treeNode* temp=*root,*p1=NULL;
    while(temp)
       {
	       p1=temp;
       	  if(temp->value>value)
       	    temp=temp->left;
       	  else
       	    temp=temp->right;
       }
    if(p1->value>value)
       p1->left=toBe;
    else
       p1->right=toBe;
    toBe->parent=p1;

 }

//此处是相关的代码
<strong>bool isSameTree(treeNode* root1,treeNode*root2)
{
	if(root2==NULL&&root1==NULL)//两个都等于NULL
	    return true;
	if(root1==NULL||root2==NULL)//只有一个等于NULL
	  return false;
    if(root1->value==root2->value)
    return isSameTree(root1->left,root2->left)&&isSameTree(root1->right,root2->right);
    else
    return false;
}
bool  isSubTree(treeNode* root1,treeNode *root2)
{  bool isSub=false;
    if(root1!=NULL)
   {
     if(root1->value==root2->value)
       {
       	isSub=isSameTree(root1,root2);
       }
       if(!isSub)
	   isSub=isSubTree(root1->left,root2);
	   if(!isSub)
         isSub=isSubTree(root1->right,root2);
   }
return isSub;
}
</strong>
int getHeight(treeNode* root)
{
	if(root==NULL)
	   return 0;
	return getHeight(root->left)>getHeight(root->right)?
	(getHeight(root->left)+1):(getHeight(root->right)+1);

}
void  printByLevel(treeNode *root,int level)
{
  if(root==NULL||level<0)
  return ;
  else if  (level==0)
    {
    	cout<<root->value<<" ";
    	return ;
    }

   else
  {
  	printByLevel(root->left,level-1);
	  printByLevel(root->right,level-1);
  }  

}
int main()
{
    treeNode* root=NULL,*root2=NULL;
    insert(&root,20);
    insert(&root,10);
    insert(&root,30);
    insert(&root,15);
    insert(&root,25);
	insert(&root,35);
    insert(&root,5);
    insert(&root2,10);
    insert(&root2,5);
    insert(&root2,100);
    int high=getHeight(root);
    for(int i=0;i!=high;i++)
      {
      	printByLevel(root,i);
      	   cout<<endl;
		  cout<<"---------"<<endl;
      }
    int high2=getHeight(root2);
    for(int i=0;i!=high2;i++)
      {
      	printByLevel(root2,i);
      	cout<<endl;
		  cout<<"---------"<<endl;
      }
      cout<<isSubTree(root,root2);
}

时间复杂度分析，假设root1的节点数为m.

<span style="font-family: Arial, Helvetica, sans-serif;"> </span><span style="font-family: Arial, Helvetica, sans-serif;">root2的节点数为n.则最坏时间复杂度为O（m*n）</span>

<span style="font-family:Arial, Helvetica, sans-serif;">算法的步骤是这样子的：</span>

<span style="font-family:Arial, Helvetica, sans-serif;">遍历tree1 </span>

找与tree2 根节点相等的节点p2

每次找到p2.就检查以p2为根节点的子树是否与tree2相等，如果相等，就停止遍历

假设tree1 有k个节点值与tree1的根节点相等 则实际最坏时间复杂度为 O(m+k*n)

因而 时间效率还是可以接受的

<span style="font-family: Arial, Helvetica, sans-serif;"> </span>