题解 | #树的子结构#

思路

1、递归
2、辅助函数
3、先找父节点的值相等的节点，进行IsPart判断；如果结果为false，那么对左右节点分别递归调用函数。
4、注意边界条件、递归终止条件

代码

/*
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};*/
class Solution {
public:
    bool IsPart(TreeNode* tree1, TreeNode* tree2){

        if(tree2==nullptr){
            return true;
        }

        if(tree1==nullptr){
            return false;
        }
        if(tree1->val!=tree2->val){
            return false;
        }

        return IsPart(tree1->left, tree2->left) && IsPart(tree1->right, tree2->right);



    }

    bool HasSubtree(TreeNode* pRoot1, TreeNode* pRoot2) {
        bool found = false;
        /*
        if(pRoot1!=nullptr && pRoot2!=nullptr){
            if(pRoot1->val==pRoot2->val){
                found = IsPart(pRoot1, pRoot2);
            }

            if(found==false){
                found = HasSubtree(pRoot1->left,  pRoot2);
            }
            if(found==false){
                found = HasSubtree(pRoot1->right, pRoot2);
            }
        }
        */

        if(pRoot1!=nullptr && pRoot2!=nullptr){
//             if(!found){
//                 found = IsPart(pRoot1, pRoot2);
//             }
//             if(!found){
//                 found = IsPart(pRoot1->left, pRoot2);
//             }
//             if(!found){
//                 found = IsPart(pRoot1->right, pRoot2);
//             }

            if(pRoot1->val==pRoot2->val){
                found = IsPart(pRoot1,pRoot2);
            }
            if(!found){
                found = HasSubtree(pRoot1->left, pRoot2);
            }
            if(!found){
                found = HasSubtree(pRoot1->right, pRoot2);
            }
        }

        return found;



    }
};