LintCode 子树

容易 子树

19%

通过

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。

您在真实的面试中是否遇到过这个题？ 

Yes

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

/** * 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) {         bool result  = false;        if (T2 == nullptr) {            return true;        }        if (T1 == nullptr) {            return false;        }        // write your code here        if (T1->val == T2->val) {             result = dp(T1,T2);        }        if (!result) {          result =  isSubtree(T1->left,T2);        }        if (!result) {            result =  isSubtree(T1->right,T2);        }        return result;    }        bool dp (TreeNode *T1, TreeNode *T2) {            if (T1 != nullptr && T2!=nullptr && T1->val == T2->val) {            return dp(T1->left,T2->left) && dp (T1->right,T2->right);        }        if (T1 == nullptr && T2 == nullptr) {            return true;        }        return false;    }};