树的子结构

面试题目：输入两颗二叉树A，B，判断B是不是A的子结构；

#include <iostream>
#include <stack>
using namespace std;

typedef struct BinaryTreeNode{
	int value;
	BinaryTreeNode * lchild;
	BinaryTreeNode *rchild;
}BinaryTreeNode;
typedef BinaryTreeNode * BiTree;

void CreateBiTreeByLevel(BiTree &t,int Array[],int i,int len)
{
	if(Array[i]==0||i > len) return;
	t = new BinaryTreeNode;
	t->value = Array[i];
	t->lchild = NULL;
	t->rchild = NULL;
	CreateBiTreeByLevel(t->lchild,Array,2*i,len);
	CreateBiTreeByLevel(t->rchild,Array,2*i+1,len);
}

void display(BiTree *t)        //显示树形结构 
{
	if(*t!=NULL)
	{
		cout<<(*t)->value;
		if((*t)->lchild!=NULL)
		{
			cout<<'(';
			display(&(*t)->lchild);
		}
		if((*t)->rchild!=NULL)
		{
			cout<<',';
			display(&(*t)->rchild);
			cout<<')';
		}
	}
}
bool ifTheSameTree(BiTree &t1,BiTree &t2)
{
	if (t2 == NULL)return true;//B树为空，肯定是相同的树
	if(t1 == NULL)return false;
	if (t1->value!=t2->value)return false;//节点不同，不是
	else
	{
		return ifTheSameTree(t1->lchild,t2->lchild) & ifTheSameTree(t1->rchild,t2->rchild);//左右均为true返回true
	}
}
bool isSubBinTree(BiTree &t1,BiTree &t2)
{
	bool result = false;
	if(t1 != NULL && t2 != NULL)
	{
		if (t1->value == t2->value)//根节点相同，就判断A，B树是否一样
		{
			result = ifTheSameTree(t1,t2);
		}
		if(!result)result = isSubBinTree(t1->lchild,t2);//判断左子树
		if (!result) result = isSubBinTree(t1->rchild,t2);//判断右子树
	}
	return result;
}

int main()
{
    BiTree A,B;
    //InitBiTree(&T);
    int a[14] = {0,1,2,3,4,5,6,0,0,0,7,0,8,9};//从下标1开始，0标识没有节点
	int b[4] = {0,6,8,9};
	CreateBiTreeByLevel(A,a,1,13);
	CreateBiTreeByLevel(B,b,1,2);
	display(&A);
	display(&B);
	if(isSubBinTree(A,B))cout<<"B is A subTree";
	else cout<<"B is Not A subTree";
	system("pause");
}

来源：http://blog.csdn.net/xiaobo620/article/details/7957539