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 HasSubtreeCore(TreeNode* pRoot1, TreeNode* pRoot2) {
        if (pRoot2 == nullptr)  return true;  //先判断2后判断1
        if (pRoot1 == nullptr) return false;
        if (pRoot1->val == pRoot2->val)
            return HasSubtreeCore(pRoot1->left, pRoot2->left) &&
                   HasSubtreeCore(pRoot1->right, pRoot2->right);
        else
            return false;
    }
    bool HasSubtree(TreeNode* pRoot1, TreeNode* pRoot2) {
        if (pRoot1 == nullptr || pRoot2 == nullptr) return false;
        return HasSubtree(pRoot1->left, pRoot2) ||
               HasSubtree(pRoot1->right, pRoot2) ||
               HasSubtreeCore(pRoot1, pRoot2);

    }
};