LintCode 子树

easy 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例子以下的样例中 T2 是 T1 的子树：

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

以下的样例中 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 hereif (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;}};