剑指offer 18：树的子结构

/*
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};*/
class Solution {
public:
    bool HasSubtree(TreeNode* pRoot1, TreeNode* pRoot2)
    {       bool isSub=false;
           //出现两树中有任意一个为空，则返回false
           if(pRoot1!=NULL && pRoot2!=NULL){
               if(pRoot1->val==pRoot2->val){
                   isSub=doesTree1HaveTree2(pRoot1,pRoot2);
               }if(!isSub){
                   isSub=HasSubtree(pRoot1->left,pRoot2);
               }if(!isSub){
                  isSub=HasSubtree(pRoot1->right,pRoot2); 
               }
           }
         return isSub;
    }
    bool doesTree1HaveTree2(TreeNode* pRoot1, TreeNode* pRoot2){
        if(pRoot2==NULL){
            return true;
        }
        if(pRoot1==NULL){
            return false;
        }
        if(pRoot1->val!=pRoot2->val){
            return false;
        }
        return doesTree1HaveTree2(pRoot1->left,pRoot2->left) && doesTree1HaveTree2(pRoot1->right,pRoot2->right);
    }
};