剑指 Offer 26. 树的子结构

class Solution {
public:
   // 仍然是前序遍历，逐渐减少 子树的高度即可，
    bool isSubStructure(TreeNode* A, TreeNode* B) {
        if(A == nullptr || B == nullptr) return false;
        return recur(A, B) || isSubStructure(A->left, B) || isSubStructure(A->right, B);
    }
private:
    // 判定条件
    bool recur(TreeNode *A, TreeNode *B) {
        if(B == nullptr) return true;
        if(A == nullptr || A->val != B->val) return false;
        return recur(A->left, B->left) && recur(A->right, B->right);
    }
};