LintCode 子树

有两个不同大小的二进制树： T1 有上百万的节点； T2 有好几百的节点。请设计一种算法，判定 T2 是否为 T1的子树。

样例

下面的例子中 T2 是 T1 的子树：

       1                3
      /               / 
T1 = 2   3      T2 =  4
        /
       4

下面的例子中 T2 不是 T1 的子树：

       1               3
      /                
T1 = 2   3       T2 =    4
        /
       4

分析：醉了！！！

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */
class Solution {
public:
    /**
     * @param T1, T2: The roots of binary tree.
     * @return: True if T2 is a subtree of T1, or false.
     */
    bool isSubtree(TreeNode *T1, TreeNode *T2) {
        // write your code here
        bool result=false;
        if(T2 == nullptr)
        return true;
        if(T1 == nullptr)
        return false;
   
      if(T1->val == T2->val)
      {
          result=DoesTree1HaveTree2(T1,T2);
      }
      if(!result)
      {
          result=isSubtree(T1->left,T2);
          
      }
      if(!result)
      {
          result=isSubtree(T1->right,T2);
      }
      
      return result;
    }
    
    bool DoesTree1HaveTree2(TreeNode *T1, TreeNode *T2)
    {
         if (T1!=nullptr && T2!=nullptr && T1->val == T2->val)
         return DoesTree1HaveTree2(T1->left,T2->left) && DoesTree1HaveTree2(T1->right,T2->right);//刚开始一直错误 原来是这块的问题 必须要写T1不为NULL和T2不为NULL
         
        if (T1 == nullptr && T2 == nullptr) 
        {
            return true;
        }
        
        return false;
         
    }
};