17. 树的子结构

https://www.nowcoder.com/practice/6e196c44c7004d15b1610b9afca8bd88?tpId=13

题目描述

输入两棵二叉树A，B，判断B是不是A的子结构。（ps：我们约定空树不是任意一个树的子结构）

输入

{8,8,#,9,#,2,#,5},{8,9,#,2}

返回值

true

/*
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||pRoot2== nullptr) {
            return false;
        }
        if (pRoot1 == pRoot2) {
            return true;
        }
        bool res = false;
        if (pRoot1->val == pRoot2->val) {
            res = isSubTree(pRoot1, pRoot2);
        }
        if (res) {
            return true;
        }
        return isSubTree(pRoot1->left, pRoot2) || isSubTree(pRoot1->right, pRoot2);
    };
    bool isSubTree(TreeNode *pRoot1, TreeNode *pRoot2) {
        if (pRoot1 == pRoot2) {
            return true;
        }
        if (pRoot2 == nullptr) {
            return true;
        }
        if (pRoot1 == nullptr) {
            return false;
        }
        if (pRoot1->val != pRoot2->val) {
            return false;
        } else {
            return isSubTree(pRoot1->left, pRoot2->left) && isSubTree(pRoot1->right, pRoot2->right);
        }
    }
};