题解 | #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) {
        if (pRoot1 == nullptr) return false;
        if (pRoot2 == nullptr) return false;
        bool r = false;
        if(pRoot1 != nullptr && pRoot2 != nullptr){
            if(pRoot1->val == pRoot2->val){
                r =  DoTree1HasTree2(pRoot1, pRoot2);
            }
            if(!r){
                r = HasSubtree(pRoot1->left, pRoot2);
            }
            if(!r){
                r = HasSubtree(pRoot1->right, pRoot2);
            }
        }
        return r;
    }
    bool DoTree1HasTree2(TreeNode* pRoot1, TreeNode* pRoot2){
        if(pRoot2 == nullptr) return true;
        if (pRoot1 == nullptr) return false;
        if(pRoot1->val != pRoot2->val) return false;
        return DoTree1HasTree2(pRoot1->left, pRoot2->left) && DoTree1HasTree2(pRoot1->right, pRoot2->right);
    }
};