树的子结构

输入两颗二叉树A，B，判断B是不是A的子结构。

遍历A，然后用A的每个节点与B进行判断

/*
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 res=false;
        if(pRoot1!=NULL&&pRoot2!=NULL){
            if(pRoot1->val==pRoot2->val)
                res=isSubtree(pRoot1,pRoot2);
            if(!res)
                res=HasSubtree(pRoot1->left,pRoot2);
            if(!res)
                res=HasSubtree(pRoot1->right,pRoot2);
        }
        return res;
    }
private:
    bool isSubtree(TreeNode* pRoot1, TreeNode* pRoot2){
        if(pRoot2==NULL) return true;
        if(pRoot1==NULL) return false;
        if(pRoot1->val==pRoot2->val)
            return isSubtree(pRoot1->left,pRoot2->left)&&isSubtree(pRoot1->right,pRoot2->right);
        return false;
    }
};