# Leetcode : 0572. Subtree of Another Tree (Tree) ###### tags:`leetcode` ``` > DFS method /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */ class Solution { public: bool isSubtree(TreeNode* root, TreeNode* subRoot) { if(!root) return false; if(isSametree(root,subRoot)){ //if root same subRoot return true; }else{ return isSubtree(root->left,subRoot) || isSubtree(root->right,subRoot); // check root->left and root->right. } } bool isSametree(TreeNode* root, TreeNode* subRoot){ if(!root || !subRoot){//check below is same return root==NULL && subRoot==NULL; }else if(root->val == subRoot->val){ //check val. return isSametree(root->left,subRoot->left) && isSametree(root->right,subRoot->right); }else{ return false; } } }; ``` >Tip condtion >first , root ->left and right last is null ,euqal subRoot is true; >second , val equal val >third , if root not same , find next on root->left , root->right.